/J^'"^- University of California • Berkeley The Theodore P. Hill Collection Early American Mathematics Books / , -^-^2a/^ ^^^l 94 > / ^ / 4 > . WRITTEN ARITHMETIC, COMMON AND HIGHER SCHOOLS; TO WHICH IS ADAPTED A COMPLETE SYSTEM OF REVIEWS, Ilf THE FORM OF DICTATION EXERCISES. BY G. A. WALTON, PRINCIPAL OF OLIVER GRAMMAR SCHOOL, LAWRENCE, MASS. BOSTON: 1868. Entered, according? to Act of Confrrcss, in the year 1804, by G. A. WALTON, In the Clerk's Office of the District Court of tlie District of Massachusetts. ELECTROTYPED AT THE BOSTON STEREOTYPE FOUNDRY, 4 SPRING LANE. PREFACE. This book is designed to prepare the pupil to meet the demands of actual life. It is itself copious in examples of a great variety of forms, and largely of a practical character ; and the accompanying Key con- tains a set of Dictation Exercises, adapted to every important topic treated in the book, to be used at the discretion of the teacher, by means of which the amoimt of practice may be increased almost indef- initely. All that the book contains is written for the pupil j and if he will learn it understandingly, he may master the principles of arithmetic with but little aid from the teacher. In the arrangement of subjects, that order has been adopted which experience has shown to be the best for all classes of learners. Some subjects, of little importance, have been briefly treated ; others have been transferred to the Appendix. Should any subject, as Duodeci- mals, Circulating Decimals, or Average of Accounts, or any examples prove too difiicult for the younger scholar, they can be omitted till the book is reviewed. Answers are given to the examples, so far as is necessary to assure the pupil that he understands the principles ; but every important prin- ciple is likewise tested by examples having no answers in the book. The answers not contained in the book may be found in the Key, from which they can easily be transferred to the black-board, if the teacher prefers to have them placed before his pupils. To determine the adaptation of a text-book to school purposes, it (3) 4 PREFACE. must be used in the school-room. This treatise has already been suc- cessfully tested by this standard, since its general character has been determined by the actual demands of a large grammar school, at pres- ent and for several years in the charge of the undersigned, and since it is largely illustrated by examples which have been repeatedly em- ployed to familiarize students with the principles they here exempHfy. Its practical character is fully certified by the testimony of many of its students, now business men, who practise its methods in the office and in the counting-room. Though, at the request of the publishers, but one name appears upon the title-page as author, the book is the joint production of the person whose name it bears, and of E. N. L. Walton, former teacher in one of the State Normal Schools of Massachusetts ; and whatever mer- its or defects the book may be found to possess, may be attributed equally to each. Our grateful acknowledgments are due to many teachers and business men for valuable suggestions, particularly to Wm. J. Rolfe, A.M., of Cambridge, for important criticisms while the work was in prepara- tion for the press ; to Francis Cogswell, Esq., of Cambridge, for hints on methods of Reviews ; and to the teachers of the Oliver Grammar School, Lawrence, for their kind assistance in solving and testing examples. GEO. A. WALTON. Lawrence, Oct. 1, 1864. TABLE OF CONTENTS. SIMPLE NUMBERS. Page Definitions, 9 Notation and Numeration, 9 Roman Notation, 10 Arabic Notation, 12 Numeration Table, . 14 Fractional Notation, 18 Addition, 19 Page Table for Practice, 22 Subtraction, 25 General Review, No. 1, 29 Multiplication, 30 Division, 3d Questions for Review, 41 Miscellaneous Examples, 42 FEDERAL MONEY. Table for Federal Money, 45 i Fundamental Operations, 46 Reduction, 46 | Bills, 49 Analysis, 53, 249, 254 Questions for Review 54 | General Review, No. 2, 65 PROPERTIES OF NUMBERS. Definitions, 56 Divisibility of Numbers, 57 Table of Prime Numbers, 59 Factoring Numbers, . . , Greatest Common Divisor, 61 FRACTIONS. Definitions, 64 General Principles, 65 Reduction to liOwest Terms, .... 6? Cancellation, . 68 Reduction of Whole and Mixed Num- bers to Improper Fractions, .... 70 Reduction of Improper Fractions to "Whole or Mixed Numbers, .... 70 Multiplication of Fractions, 72 Reduction of Compound Fractions, . 73 Division of Fractions, 74 Reduction of Complex Fractions, . . 77 To find the Whole from a Part, ... 78 What Part one Number is of another, 80 Least Common Multiple, 81 Common Denominator, 84 Addition of Fractions, 85 Subtraction of Fractions, 86 Greatest Common Divisor of Frac- tions, 88 Least Common Multiple of Fractions, 88 Questions for Review, 89 Miscellaneous Examples, ...... 91 General Review, No. 3, 97 (V) VI TABLE OF CONTENTS. COMPOUND DENOMINA.TE NUMBERS. Page Definitions, 98 Federal Money, 9S English Money, 99 Reduction, 100 Comparison of English and Federal Money, 101 Troy Weight, 102 Apothecaries' Weight, 103 Avoirdupois Weight, 104 Comparison of Weights, 105 Long Measure, 105 Surveyor's Measure, 107 Mariner's Measure, 107 Cloth Measure, 108 Square Measure, 108 Cubic Measure, Ill Liquid Measure, 113 Dry Measure, 113 Page Comparison of Liquid and Dry Meas- ures, lit Circular Measure, 114 Time Measure, 116 3Iiscellaneous Table, 119 Reduction of Fractions to Whole Numbers of Lower Denominations, 120 Reduction of Whole Numbers to the Fraction of a Higher Denomina- tion, 121 Compound Addition, 123 Compound Subtraction, 126 Table of Latitudes and Longitudes, . 130 Compound Multiplication, 132 Compound Division, 133 Longitude and Time, 134 Questions for Review, 135 Miscellaneous Examples, 138 Duodecimals, 145 | General Review, No. 4, 149 DECIMAL FRACTIONS. Definitions, 150 Numeration Table, 150 Addition, 153 Subtraction, 154 Multiplication and Division by 10, 100, &c., 155 Multiplication, 156 Division, 157 Reduction, 159 Circulating Decimals, 160 Reduction of Compound Numbers to Decimals of a Higher Denomina- tion, 162 Reduction to Whole Numbers of • Lower Denominations, 163 Questions for Review, 164 Miscellaneous Examples, 165 Peactice, 166 I General Review, No. 5, PERCENTAGE. Definitions and Reductions, 171 To find any Per Cent, of a Number, . 172 To find 100 Per Cent, from a given Per Cent., 174 To find what Per Cent, one Number is of another, 175 Profit and Loss, 176 Interest, 178 Partial Payments, 187 Annual Interest, 194 Compound Interest, 195 Problems in Interest, 199 Present Worth and Discount, .... 202 Banking, 201 General Review, No. 6, . 208 Commission, Brokerage, and Stocks, 209 TABLE OF CONTENTS. vu PERCENTAGE. — (Continued.) I»surance, Page . 213 Exchanjrc, Average or Equation of Payments, . 215 Average of Accounts 221 Taxes, 225 Custom House Business, 228 Pago .230 Questions for Review, 236 Miscellaneous Examples, 238 General Review, No. 7, . 244 MISCELLANEOUS. Ratio, 246 Proportion, 248 Analysis and Simple Proportion, . . 249 Analysis and Compound Proportion, 254 Simple Partnership, 258 Compound Partnership, 200 Questions for Review, 203 Involution, 204 Evolution, 205 Square Root, 206 Application of Square Root, 272 Cube Root, 276 Mensuration, 283 Circles, Similar Triangles, Polygons, 293 Similar Solids, 295 Questions for Review, 297 General Review, No. 8, 299 Alligation 3Iedial, 300 Alligation Alternate, 302 Arithmetical Progression, 305 Geometrical Progression, 309 Annuities, 311 Questions for Review, 314 Miscellaneous Examples, 315 APPENDIX. Divisibility by 9, 325 Proof of Multiplication and Division by casting out tlie 9's, 325 Contractions in Multiplication, . . . 320 Contractions in Division, 828 Exact Interest by Days, 329 Methods of Computing Time, ... .329 Table for finding Diiference of Days 331 Mensuration of Timber, &c., . . . .332 Gauging, 333 Interest at 7 3-10^, a35 Metric System, 336 PUBLISHERS' NOTICE. WALTON'S ARITHMETICS. THE SERIES CONSISTS OF THREE ROOKS, VIZ. : I. Tlie rictor-ial Primary -A^ritliinetlc. II. Tlie Intellectual A.ritliinetlc. III. Tlie TV^ritten .A.r'itliiiietic. The publishers invite the attention of Teachers and School Officers to this series of Text-Books, confident that on examination they will commend themselves to every practical educator. No other series, in general use, with which they are acquainted, comprises a full course of Arithmetic in Three Books. WALTON'S DICTATION EXERCISES are supplementary to AValton's Series, and afford a large amount of practice in the fundamental rules, and in all the important prac- tical applications of arithmetic. They are designed for reviews and test exercises, and may be used at any stage of the pupil's progress, and in connection with any series of arithmetics. Course of Study. I. Complete the Primary Arithmetic before commencing the Intellectual. II. Complete the Intellectual Arithmetic to page 63, together with written exercises [see foot-notes of Intellectual], before com- mencing the Written Arithmetic. III. Continue the study of the Intellectual in connection with tlic Written Arithmetic, observing to complete the subjects in the Intel- lectual before commencing them in the Written. AEITHMETIC. Article 1, Arithmetic is the science of numbers, and the art of computing by them. S, A Unit is one. S. A Number is a unit, or a combination of units. 4, A Concrete, or Denominate Number, is a number which is applied to some object or objects ; as, one hoy, two apples f three slate pencils, four sounds. 5, An Abstract Number is a number which is not applied to any object ; as, one, two, three. 6. Exercise. Name the concrete numbers in the following list : — Four girls ; seven swans ; two ; ten ; nine chairs ; five knives ; eight ; twelve horses ; six mules ; two oxen ; four ; eleven ; seven pond lilies ; one ; ten ; thirteen ; nine days ; fifteen lessons ; two rabbits ; six bushels. Name the abstract numbers in the above. T. The fundamental operations of written Arithmetic are based upon Notation, and consist of Addition, Subtrac- tion, Multiplication, and Division. NOTATION AND NUMERATION. 8. Notation is the art of writing numbers. Numeration is the art of reading numbers. O. Besides being expressed in words, numbers are repre* (9) 10 SIMPLE NUMBERS. sented by letters and JigKres. Tlic method of representing them by letters is called the Roman method, because it was used by the ancient Romans. The method of representing them by figures is called the Arabic method, because our first knowledge of it was obtained from the Arabs. Roman Method. 10, The Roman Method is principally used in writing dates, and in numbering chapters and sections of books. 11, It employs seven capital letters; I representing one; Y, five ; X, ten ; L, fifty ; C, one hundred ; D, five hundred ; M, one thousand. 13, By combining these letters in various ways, all num- bers may be expressed, the following principles being ob- served : — (1.) When a letter is repented, its value is repeated, (2.) When a letter is placed before another of greater value, its value is to be taken from that of the greater; thus, IV denotes four. (3.) AVhen a letter is placed after another of greater value, its value is to be added to that of the greater; thus, VI de- notes six. (4) AVhen a letter is placed between two of greater value, its value is to be tahen from their united value ; thus, XIX denotes nineteen. (5.) Any letter may be made to express thousands instead of units by placing a dash over it. Thus X denotes ten thousand ; B, five hundred thousand ; M, one thousand thousand, or one million. 13, Table of Roman Notation. denotes five, six. seven, eight. ♦ mi is sometimes used for four. I denotes one. V II two. VI ni three. VII IV* four. VIII NOTATION AND NUMERATION. 11 IX * denotes nine. L denotes fifty. X ten. LX sixty. XI eleven. LXX seventy. XII twelve. LXXX eighty. XIII thirteen. XC ninety. XIV fourteen. C one hundred. XV fifteen. CC two hundred. XVI sixteen. CCC three hundred. XVII seventeen. CD four hundred. XVIII eighteen. D five hundred. XIX nineteen. DC . six hundred. XX twenty. DCC seven hundred. XXI twenty-one. DCCC eight hundred. XXX thirty. CM nine hundred. XXXI thirty-one. M one thousand. XL forty. M one million. XLI forty-one. MM two million. 14. Exercises. Read or write in words the following numbers: — 1. IV. 2. XL 3. XIV. 4. XIX. 5. XXVL 6. XXIX. 7. XXXVL 8. XL. 9. XLV. 10. XLIX. 11. Lvm. 12. LXXL 13. LXXXIX. 14. XCVIII. 15. CLV. 16. CXIX. 17. CCCXLVIL 18. CDLXXIL 19. DCCXLIV. 20. MDXCIV. 21. MDCCCLXIV. 22. MD. 15. VTrite the following in Roman characters : — 1. All the numbers from one to twenty, inclusive. 2. All the numbers from thirty to forty, inclusive. 3. All the numbers from ninety to one hundred, inclusive. 4. One hundred thirty-eight. * yill| is sometiines used for nine. 12 SIMPLE NUMBERS. 5. Three hundred twenty-four. 6. Four hundred forty-nine. 7. Five hundred eighty-six. 8. Seven hundi-ed sixty-seven. 9. Nine hundred fifty-three. 10. One thousand four hundred seven. _ 11. Five thousand eight hundred. Ans. VDCCC. 12. Ten thousand ninety-nine. 13. One thousand eight hundi-ed sixty-four. Akabic Method. 16, The Arabic Method of representing numbers employs ten characters, or figures, as follows : — 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. One, Two, Thi-ee, Four, Five, Six, Seven, Eight, Nine, Zero. 17, The first nine are called digits, from the Latin word digitus, ajlnger, it being supposed that the ancients first counted by their fingers. They are also called signijicant figures, be- cause they are signs for numbers. The character, 0, called zero, signifies nothing when it stands alone. It is called a figure of place because, in writing numbers, it is used to fill places not occupied by other figures. Used singly, these characters can represent only the numbers from one to nine ; but combined according to the following prin- ciples, they are used to represent all numbers. 18, The figures which represent simple units are placed at the left of a dot, called the decimal point. (See Art. 23 and note.) The first place, therefore, at the left of the decimal point is called the units' place ; thus, 7. is read " seven units," or " seven." Having no single figure to represent ten units, we consider the collection of ten units as one ten, or a unit of the second order^ and represent it by the figure 1 put in the tens' place, which is the second place from the decimal point towards the left ; thus, 10. represents ten, the zero being used to fill the units' place, which would otherwise be vacant. If we have any number of NOTATION AND NUMERATION. 13 tens and units to write together, we put the number of tens in the tens' place, and of units in the units' place; thus, thirty-six, or three tens and six units, is written 36. A collection of ten tens is called one hundred, or a unit of the third order, and is represented by 1 in the third or hundreds' place, (100.) ; two hundreds are represented by 2 in the hundreds' place, (200.) ; three hundreds by 3 in the hundreds' place, (300.) ; etc. A collection of ten hundreds is called one thousand, or, a unit of the fourth order, and is represented by 1 in the fourth or thousands' place (1000.) ; two thousands are represented ' y 2 in the thousands' place (2000.) ; etc. i § § c 1 H a H p Q 7632. The above represents seven thousands, six hundreds, three tens, and two units, and is read, " Seven thousand six hundred thirty^ two." Exercises. Read the followino: : — 19. 1, 2. 3. 4. 5. 20. 86. 132. 6321. 7862. 99. 7. 10. Write in figures, — 1. One thousand six hundred forty-four. 2. Two thousand eight hundred twenty-one. 3. Nine hundred nine. 4. Six thousand two hundred ten. 5. Eight thousand eight. 6. Five thousand fifty. 7. Seven thousand seven hundred seventy. 8. Twenty-nine. 9. Six hundred two. 10. Six thousand twenty. 428. 11. 9000. 1302. 12. 9090. 6006. 13. 9009. 7801. 14. 8047. 541. 14 / SIMPLE NUMBERS. Questions. — What is the first place at the left of the decimal point called ? What is the second place at the left called ? The third ? How many units make one ten? How many tens make one hundred? How many hundreds make one thousand? How many units make one hundred? How many units make one thousand? How many tens make one thousand? What are units of the first order called? Ans. Simple units. What are units of the second order called? Of the fourth ? Of the third ? In 7632 how many tens, and what number remains ? How many hundreds, and what remains ? How many thousands ? Hkmakk. — The number of units of any order is sometimes called a term ; thus, the terms of 632 are 6 hundreds, 3 tens, and 2 units. SI, Numeration Table. CO CO O .2 . . '^ i3 :=3 • 2 <» 52 S 1 s -^ I I .1 I i4 '4.J ^-I'i-I^-) 4->4>> .4-34J4-> .t_)4->-«^ 4-l4_'4-> r-IO05 COl-O "O'^CO (M»-lO 05G0 1- CO'OTf^ *2 8 4, 9 6 ?, 3 4 r, 2 oT, 8 7 G, 3 2 2, 1 2 4 TID 'CC ~»o "3. fOjr tSriS oo oo og o* o2 Oq 'Si 'E i 1:2 i .2 -go -C rt f^S P^^ e^j= f^= fSs ;i:;s p,c 5-5 £« £h 5M S^ sI t3 S^3* The ^i^/i place from the decimal point towards the left IS the ten thousands^ place, each ten-thousand being equal to ten of the thousands ; iha' sixth place is the hundred thousands' place, each hundred thousand being equal to ten ten-lhousands ; and so on, each' unit of any order being equal to ten units of the order immediately preceding. We now see that the number of units of any order is expressed by the figure^ and the order of units by the flace which the figure occupies ; or, in other words, the value represented by any NOTATION AND NUMERATION. 15 figure depends upon the figure itself, and upon the place which that figure occupies. Thus, 2 in the first place means simply two (that is, two units) ; in the second place, it means two tens, or twenty ; in the third place, two hundreds. S3. Since, by this method of writing numbers, the value represented by a significant figure increases as that figure is re- moved towards the left, and decreases as it is. removed towards llie right, by a scale of tens, the system is called the Decimal System, from the Latin word decern, which signifies ten. Note. — The reason for calling the dot (Art. 18) a decimal ijoint must now bo obvious. This point is not always written, but, when not writ- ten, it is always understood. 24. By examining the table (Art. 21), we find it sep- arated by commas into groups of three places each. These j]froups are called periods, the first period being that of units ; the second that of thousands ; the third, millions ; the fourth, billions, etc. Thus we have simple units,, tens of units, and hundreds of units ; units, tens, and hundreds of thousands ; units, tens, and hundreds of millions ; etc. 52#>, Exercises ox the Table. 1. Give the names of the first two periods from the decimal point, reading them towards the left ; towards the right. Give the names of the first three periods in the same way ; of the first four ; five ; six ; seven. What is the second period called ? third ? sixth ^ seventh ? fourth? fifth? 2. In which period are found thousands ? millions ? simple units ? trillions ? billions ? quintillions ? quadrillions ? 3. In which place of what period are found tens of units ? thou- sands ? hundred-thousands ? millions ? hundreds of units ? ten- thousands ? billions ? hundred-millions ? ten-billions ? ten-millions ? quadrilUons ? ten-quintillions ? hundred-biUions ? hundred-trillions ? quintillions ? ten-quadrillions ? ten-trillions ? hundred-quadrillions ? trillions ? hundred-quintillions ? 4. Name the order of units of each number in paragraph 3. Ans, Tens are of the second order, thousands of the fourth order ; etc. 5. "What order of units is found in the first place of the second period ? Ans. Fourth order, or thousands. In the third place of the 16 SIMPLE NUMBERS. first period ? In the second place of the third period ? In the third place of th 3 fourth period ? In the first j^lace of the fifth period ? In the third place of the sixth period? In the second place of the seventh period ? In the third place of the third period ? In the first place of the seventh period ? In the second place of the fourth pe- riod ? In the first place of the sixth period ? 6. In 6480921 how many tens, and what remains ? Ans. G48092 tenSj and 1 unit remaining. How many hundreds, and what remains ? Ans. 64809 hundreds, and 21 remaining. How many millions, and what remains ? thousands ? ten-thousands ? hundred-thousands ? 2G. The names of the periods employed to express numbers higher than Quintillions are, in their order from Quintillions, Sextillions, Septillions, Octillions, Nonillions, Decillions, Unde- cillions, Duodecillions, Tredecillions, Quatuordecillions, Quinde- cillions, Sexdecillions, Septendecillions, Octodecillions, Novende- cillions, Vigintillions, etc. 37. To read numbers, observe the following Rule. — Beginning at the umts' place,, point off the expression into periods of three figures each ; then begin at the left,, and read each period in order from left to rights giving after each, excepting the last, the name of the period. 38. Exercises. J. Read or write in words the follow i ng: — 1. 361. 13. 987654321. 2. 786. 14. 89743208. 3. 3261. 15. 1122334455. 4. 96321. 16. 3670980347. 5. 9301. 17. 9008007006. 6. 80021. 18. 12400496623. 7. 654237. 19. 245607000000. 8. 9326429. 20. 94632748632. 9. 9000200. 21. 1781006390800. 10. 86320029. 22. 62876432019623 11. 324867. 23. 753248734762869. 12. 81402020. 24. 943300896402798. See Dictation Exev'^ises, Key. NOTATION AND NUMERATION. 17 do. Name the terms in the first example above, commencing with units (Art. 20, Remark). Ans. One unit, six tens, three hundreds. Name the tenns in tlie second example. In the third. In the other examples, in their order. Head from the Table (Art. 21), the number represented by the first six figures from the decimal point ; the first eight ; the first ten ; nine ; twelve J fifteen; seventeen; twenty; fourteen; eighteen. S©, To write numbers, observe the following Rule, — Beginning with the highest period, write the figures of each period in their order from left to right, filling vacant places with :::eros, 33. Exercises. Write the following : — 1. Three hundred sixty-four, Ans, 364, 2. Seven thousand eighty-nine, Ajis, 7089. 3. Eighteen thousand eighteen, 4. Nine hundred thousand sixteen. 5. Four hundred twenty thousand, six hundred eighty- three. 6. Eight hundred ten thousand, two hundred four. 7. Two hundred fifty-nine thousand, seventy. 8 Eorty-five million, seven hundred thousand, two hundred fifty- one. 9. Nine himdred one million, two hundred eighteen thousand, twenty-two. 10. Three billion, thirty-seven million, nine hundred six thousand, two hundred. 11. Two hundred thirty-four million, eight hundred sixty-three thousand, three hundred eighty-nine. 12. Seventeen billion, seven hundred fifty-nine million, ninety thou- sand, sixty-seven, 13. Three hundred thirty-three quadrillion, se\'en hundred seventy- nine billion, three hundred thousand, two. 14. Nine hundred ten quadrillion, four million, three thousand. 15. Fifty-four quintillion, eighty-three quadrillion, nine hundred million, seventeen thousand, one hundred eighty-two. 16. Eighteen billion, four. 17. Forty million, eight hundred thousand. 2 18 SIMPLE NUMBERS. 18. Eighty-n'mc million, four hundred five thousand, seven. 19. Thirty-seven trillion, ninety-three billion, eighty-one. 20. Seven hundred quintillion, one quadrillion, one. 21. Fifty quintillion, forty-nine thousand, thirty. 3S* We have seen that the value represented by a figure in- creases by a scale of tens, as the figure is removed towards the left, and decreases in the same manner as it is removed towards the right. ^ Applying this principle, we can represent parts of units by placing figures at the right of the decimal point. If we consider a unit to be composed of ten equal parts, we may represent one or more of these parts, which are called tenths, by a figure in the first place at the right of the point ; again, if we consider one of these tenths to be composed of ten equal parts, w^e may represent one or more of these parts, which are called hundredths, by a figure in the second place, and so on. The first place at the right of the point is the tenths' place, the second, the hundredths' place, the third, the thousandths' place. I il ^ IB g a E £ -S S *3 e c o Thus: .7 8 5 Here the 7 at the right of the point represents seven tenths of a whole one, the 8 represents eight hundredths, and the 5 rep- resents five thousandths. The entire number is read seven hun- dred eighty-Jive thousandths ; .25 is read twenty-jive hundredths ; .3 is read three tenths. Exercises. Read the following : — 1. .325; .763; .202; .085; .42; .6. 2. .87 ; .03 ; .504 ; .004 ; 39 ; .039., Write the following : — 1. One hundred three thousandths. Ans. .103. 2. Eight hundred twenty-one thousandths. 3. Two hundred forty-five thousandths. 4. Seven tenths. Seven hundredths. ADDITION. 19 ADDITION. 351. Addition is the process of finding a number equal in value to two or more given numbers of the same kind. The number thus obtained is called the sum, or amount. An upright cross, -f- , read plus, is the sign of addition, and. placed between t'/vo numbers, signifies that the one is to be added to the other. Two horizontal lines, r=: , read equal to, are the sign of equality, and signify that the quantities between which they are placed, are equal ; thus, 2 -[- »^ = 7, is read, two plus Jive is equal to seven, or, two plus jive equals seven. Illustrative Example. 34. Add the numbers 321, 285, and 937. Opkkation. We first write these numbers, units under units, 321 tens under tens, hundreds under hundreds, and draw 285 a line beneath. Then, adding the units first, 7 -f- 5 -f 937 1 = 13 units = 1 ten and 3 units ; we write the 3 in — ;; — the units' place, under the column of units, and An:;. 1543 reserve the 1 ten to add with the column of tens. 1 ten + 3 tens + 8 tens + 2 tens = 14 tens = 1 hun- dred and 4 tens ; we write the 4 tens in the tens' place, and reserve the 1 hundred to add with the column of hundreds. 1 Imndred + 9 hundreds-}- 2 hundi'eds -[- 3 hundreds =15 hundreds r=:l thousanfl and 5 hundreds ; we write the 5 hundreds in the hundreds' place, and the 1 thousand in the thousands' place, and thus find the amount of the given numbers to be one thousand five hundred forty-three. Hence we derive the following Rule for Addition. Write the numbers, units under units, tens under tens, hundreds under hundreds, etc. Begin to add at the units' column. If the sum of the units is less than ten, write it under the column of units ; if ten, or a number greater than ten, place the units^ f9^^^^ under the column of units, and reserve the tens to odd with the tens. Proceed in the same way with the other columns, writing down the entire amount of the last column. 20 SIMPLE NUMBERS. Proof I. — Add each column in a reverse direction ; if the same result he obtained as before, the work may he presumed to he correct. Note. — Greater readiness will be attained by mentioning only the results in adding columns. Thus, in the above example, instead of saying 7 and 5 are 12, and 1 are 13, say 7, 12, 13 ; and instead of saying, 1 ten and 3 tens are 4 tens, and 8 tens are 12 tens, and 2 tens are 14 tens, say 1, 4, 12, 14 tens. 35. Examples for Practice. 1. What is the sum of twenty-one, sixty-seven, eighty-nine, thirty-two, forty-five, thirteen, ninety, and seventy-eight ? Ans. 435. 2. What is the sum of six hundred four, nine hundred ninety- nine, seven hundred ten, six thousand nine hundred eighty-two, eleven thousand eight hundred seven ? Ans. 21,102. 3. What is the sum of 326, 981, 362, 707, 889, and 864 ? Ans. 4129. 4. What is the sum of 246, 368, 909, 896, 763, and 892 ? Ans. 4074. 5. What is the sum of 32689, 86543, 94861, 18325, and 90026? Ans. 322,444. 6. What is the sum of all the numbers from one to thirty, inclusive ? Arts. 465, 7. What is the sum of all the numbers from one hundred fifty to one hundred seventy-five, inclusive ? 8. Add 99, 364, 77, 86, 912, 32678, 96542, and 32684. 9. Add 987, 5, 679, 369, 153, 888, 806, 17, 27, and 5654. 10. Add 915, 875, 617, 868, 575, 387, 694, 946, and 6377. 11. Find the sum of the last four answers. Ans. 189,506. 12. Add 987, 425, 672, 307, 216, 321, 111,872,564,876, 318, 419, 187, 160, and 3453. 13. 875 + 466 -f 327 + 942 + 286 + 424 -f 309 + 429 + 482 + 317 + 406 + 466 + 111 + 171 + 1618 = what? 14. 324 + 868 + 522 + 297 + 789 + 524 + 286 + 361 + 472 + 884 + 472 + 287 + 649 + 592 + 1788 — what? 15. 876 + 205 + 918 + 468 + 207 + 948 + 572 + 618 + 861 + 594 + 872 + 206 + 48 -f 500 + 918 + 1331 == what? ADDITION. 21 IG. 3G196 + 5384 + 2963 + 1200 + 100200 + 2560 + 74 _|- 36 + 5 + 4786 + 186 + 544 + 396486 = what? 17. Find the sura of the last five answers. Ans. 587,294. Proof II. — Separate the example into two or more parts hy horizontal lines ; add the parts separately, and then add their amounts ; if the same result he obtained as hefore^ ike work may he presumed to he correct. See Example 18. (18.) Proof. (19.) (20.) 4163314 7137500 7984172 5949841 9345477 8194324 4956811 1233198 4221001 1726414 16796380 2122172 4754632 9876431 8914619 3241320 7325146 3141691 7987346 9136' 719 4131261 7325789 8677485 35015781 3286432 2941816 ^MS. 51812161 51812161 9710100 2861423 (21.) (22.) (23.) (24.) (25.) 449 9250 81713 247742 3482 788 19 93957 303321 6327 435 8158 38 478984 8618 663 7901 4885 98517 9532 67 6850 3750 232326 2419 455 5102 15 879416 4671 399 4372 21901 123192 8384 617 3911 86462 10921 3476 31 2514 71557 800467 2123 205 1677 99108 93219 619 871 3501 8298 63496 9600 431 5528 33984 876201 4520 219 7332 67310 23407 5418 868 9415 83568 89467 7317 189 8267 97371 77111 2982 598 6408 76503 ^ 98121 8415 529 4641 86294 267137 3618 721 2286 45939 689642 8976 256 3719 36815 232864 6521 583 5931 81541 98518 9357 SIMPLE NUMBERS. 36 > Table for Practice in the Fundamental Operations. L 21 20 19 18 ir IG 15 14 13 12 11 10 9 8 7 (5 5 4 3 2 1 A -9 8 7 4 4 9 2 5 9 8 7 9 15 3 17 1 8 4-A B-9 5 7 8 8 8 8 4 4 2 5 8 7 5 9 9 5 3 3 7-B C-6 7 9 4 3 5 5 6 8 6 7 2 6 1 7 8 7 2 6 9 2- C D -3 6 9 6 6 3 5 3 6 3 7 8 6 8 3 2 7 4 7 6-D E-1 5 3 9 6 7 8 5 6 2 16 5 7 5 4 3 6 2 8-E F-8 8 8 4 5 5 5 5 3 2 1 3 8 7 2 16 3 5 6-F G-8 6 3 9 9 5 16 111 6 9 4 5 3 5 7 5-G H-9 1 7 5 1 7 2 4 4 8 7 2 9 4 6 8 4 2 9 9 3-H 1-927 9 3 1 3 3 5 6 4 6 4 ly '4 7 1 8 8 8-^1 J -9 5 3 2 5 6 4 9 8 7 6 8 7 17 4 4 4 7-^ J K -6 6 7 8 7 1 6 6 8 3 18 2 6 8 6 2 2 6- K L -7 2 8 4 3 1 6 6 9 4 19 5.2 6 3 5 4 4 1 7- L M-8 4 5 2 19 2 4 8 1 8 7 8 4 9 5 3 4 6 2 1-M N-1 4 4 8 6 8 1 5 2 16 2 16 111 5 8 4-N 0-225 18 9 3 9 3 8 7 5 45^ 1 9 7 2 9 0- P-1 9 9 5 9 8 9 8 2 2 8 4 5 7 5 4 9 9 4 0-P Q-9 4 1 5 2 9 5 5 5 2 9 4 18 8 7 6 7 0-Q R-7 9 5 7 2 1 8 2 3 8 9 6 5 9 4 9 2 2 9 8-11 S -7 3 4 2 5 6 3 2 1 5 17 8 6 3 9 6 4 7 5- S T -3 2 3 5 8 3 5 3 8 9 4 5 16 4 8 4 6 2 7-T U-8 2 4 17 4 7 7 4 2 8 7 4 9 9 2 3 5 5 8 6-U V-7 6 9 4 16 2 6 8 4 6 3 6 4 3 8 7 2 1 9-V W-1 8 8 2 4 8 2 4 7 1 6 7 6 2 8 9 7 9 8 4-W X~8 7 2 8 9 2 9 8 2 8 7 2 8 9 8 6 2 4 7 6 8-X Y-4 4 4 7 6 4 4 2 5 5 7 4 4 5 7 4 7 6 5 1 1-Y 1* Add in the above Tiible' (as units, tens, and hundreds) 26. 1, 2, and 3. Ans, 13382. I "' ioi 13, 14, and 1^ f 27. 4, 5, and 6. \\1^'' ^l._16, 17, and 18.^ ^/ 28. 7, 8, and 9. I \^^ ^ ' i^J\ 19, 20, and 21. 29. 10, 11, and 12. J j C^?'0 33.'' 1 to 21 inclusive. 1^* For further Exercises on the Table, see Key ' H lU /'i^ ADDITION. 23 34 Paid $2400 (dollars) for my farm, $155 for my horse and cart, $26 for garden utensils, $86 for a mowing-mathine, $10 for a horse-rake, and $108 for a pair of oxen. Required the amount. 35. A body of troops were furnished with 3622 Springfield rifled muskets, 7690 smooth-bores, and 13185 Enfield rifles. Required the amount. 36. J. R. bought of the Seneca Knitting Mills, 39600 pairs of socks ; of Whitten, Hopkins & Co., 9782 pairs ; of Pierce Brothers &: Co., 9353 pairs; of Allen, Lane & Washburn, 5664 pairs; of George C. Bosson, 4296 pairs; of Cushing, Pierce & Co., 1315 pairs; of Samuel Dennis, 276 pairs. Required the amount. 37. Required the average number of pupils attending the Grammar Schools of Boston during the year 1859-60, the aver- age number attending the Adams School being 493 ; the Bigelow School, 469; Bowdoin School, 538; Boylston, 941; Brimmer, 575; Chapman, 626; Dvvight, for boys, 622; Dwight, for girls, 489; Eliot, 708; Franklin, 559; Hancock, 719; Lawrence, 761; Lincoln, 466; Lyman, 370 ; Mayhew, 367 ; Phillips, 549 ; Quincy, 720 ; Wells, 494 ; Winthrop, 933. 38. In the year 1861, Massachusetts furnished for the TJ. S. army, from her several counties, as follows : From Barnstable, 3 commissioned officers and 108 enlisted men; Berkshire, 21 offi- cers, 614 men; Bristol, 59 officers, 1681 men; Dukes, officers, 1 man; Essex, 148 officers, 4134 men; Franklin, 12 officers, 482 men ; Hampden, 35 officers, 845 men ; Hampshire, 15 officers, 575 men; Middlesex, 141 officers, 4200 men; Nantucket, 1 offi- cer, 7 men; Norfolk, 70 officers, 2031 men; Plymouth, 44 oificers, 1363 men; Suffolk, 278 officers, 4111 men; Worcester, 110 officers, 3464 men. Besides these, there joined her regi- ments, 647 men whose residences were not given, and 20 oflficers and 955 men from other States. Required the whole number of enlisted men in her regiments ; of commissioned officers ; of both. 39. Massachusetts furnished army shoes, 16649 -|- 4480 + 24 SIMPLE NUMBERS. 7139 + 3228 + 2022 + 2336 + 2220 + 1000 + 1200 + 1236 + 1013 + 240 pairs ; cavalry boots, 336 -f 1008 + 336 + 192 -|- 160 -|- 168 -|- 150 pairs. Required the number of pairs of boots; of shoes; of both. 40. She furnished hats, 12000 +4704; caps, 12130+ 2934 + 2069 + 450 + 251 + 98 + 160. Required the number of hats ; of caps. 41. On commencing business a merchant had $7752 in cash, $7719 in real estate, goods valued at $9728, a lot of cattle valued at $6930, a ship valued at $16834; during the first year he was in trade he gained above all his expenses $3195. What was he worth at the end of the year ? 42. What is the number of square miles in the British Isles, there being in Scotland 30000, in England 31200, in Wales 7200, and in Ireland 32500? 43. The United States contain 3284100 square miles more than the British Isles ; required the area of the United States? 44. What is the length of the Grand Trunk Railway from Detroit to Portland, the distance from Detroit to Stratford being 143 miles ; from Stratford to Georgetown, 59 miles ; from Georgetown to Toronto, 30 miles ; from Toronto to Coburg, 69 miles ; from Coburg to Belleville, 44 miles ; from Belleville to Kingston, 48 miles ; from Kingston to Brockville, 47 miles ; from Brockville to Prescott, 12 miles ; from Prescott to Corn- wall, 46 miles ; from Cornwall to Montreal, 67 miles ; from Montreal to Richmond, 73 miles ; from Richmond to Island Pond, 71 miles ; from Island Pond to Gorhara, 58 miles ; from Gorham to Bethel, 21 miles ; from Bethel to Danville, 42 miles ; from Danville to Portland, 28 miles ? 45. How far is it from Detroit to Toronto ? 46. How far from Toronto to Montreal ? 47. How far from Kingston to Montreal ? 48. How far from Montreal to Portland ? 49. How far from Portland to Gorham ? 50. From Boston to Portland is 111 miles ; how far is it from Boston to Montreal ? ^P For Dictation Exercises, see Key. SUBTRACTION. 25 SUBTRACTION. 37. Subtraction is the process of taking one number from anotlier of the same kind, to find the difference. The number which is subtracted is called the subtrahend, from the Latin suhtrahendus, to he taken from under ^ as that is the number taken away. The number from which the subtra- hend is taken is called the minuend, from the Latin minuendus, to he made smaller^ as that is the number to be diminished. The result is called the difference, or remainder. A short horizontal line, — , read minus or less, is the sign of subtraction, and, placed bet\veen two numbers, signifies that the number after it is to be taken from that before it ; thus, 7 — 3 :rr 4, read, seven minus three equals four, shows that, if 3 be taken from 7, the remainder is 4. Illustrative Example, I. 38. From 2G7 take 135. OPEaATioN. Tor convenience, \vc write the sub- Mmuend, 26/ trahend under the minuend, placing bubtrahend, 135 units under units, tens under tens, T> • J ion 4 hundreds under hundreds, and draw Bemamder, 132 Ans. ,. , i ^ • ,. ^ • a hne beneath ; 5 units from / units n: 2 units, which we write in the units' place, under the units ; 3 tens from 6 tens = 3 tens, which we write in the tens' place ; 1 hundred from 2 hundreds r= 1 hundred, which we write in the hun- dreds' place; and the result is 132, which is the difference between 267 and 135. 39. Proof. If 132 is the difference between 207 and 135, it is evident that, if we add 132 to 135, the sum will equal 267. Hence, to prove subtraction, add the difference to the suhtrahend. If the sum thus ohtained is equal to the minuend, the work may he presumed to he correct. Note. The pupil should prove each example, till he is suro that he makes no nristakes. 26 SIMPLE NUMBERS. 4:0. Examples. 4. 868879 — 42155 = ? 5. 974968 — 721265=:? 6. 37879868 — 1244045:1:^? 1. 368-^334 = ? ^«s. 34. 2. 2769 — 2631 = ? ^775. 138. 3. 362785-250122 = ? Sum of the last four answers = 37828933. 4L1. Illustrative Example, II. 9861— .3674 = what? Oferation. Here a difficulty presents itself. We cannot take 9861 4 units from 1 unit. In order to i)erform the opera- 3674 tion, we must reduce one of the tens in the minuend to units, which with the 1 unit we already have, Ans. G1S7 equals 11 units; 4 units from 11 units = 7 units, which we put in the units' place. Having reduced one of the tens to units, we have but 5 tens left, and as 7 tens cannot be taken from 5 tens, we must reduce one of the hundreds to tens, which r= 10 tens ; 10 tens -j- 5 tens =z 15 tens ; 7 tens from 15 tens = 8 tens, which we write in the tens' place ; 6 hundreds from 7 hundreds n^l hundred; 3 thousands from 9 thousands :zr 6 thousands, and the answer is 6187. Hence the lluLE FOR Subtraction. Wiite the subtrahend beneath the mmueiidj units under units, tens under tens, etc. Begin to sub-, tract at the units^ place, taking each term * in the subtrahend from the one above it, and placing the remainder beneath. If the upper term is less than the lower, increase it, by adding to it one of the next higher denomination reduced to its oivn denomination, and then subtract, bearing in mind, in the next operation, that the upper term has been diminished by the one reduced. Examples. What are the remainders in the following examples ? (1-) (2.) (3.) (4.) Minuend, 849 321 8642 3084 Subtrahend, 278 219 730 2427 Remainder, 571 102 7912 657 Proof, ' 849 321 ♦ See Art. 20. Remark. SUBTRACTION. 27 (5.) (6.) (7.) (8.) From 3228 3256 7862 98731 Take 409 2948 7589 19829 Sum of the last four remainders, 82,302. 9. A man had 375 oranges in a box; if he should sell 259 of them, how many would he have left ? Ans. 116 oranges. 10. A man, having 451 acres of land, gave 349 acres to his son ; what remained ? Ans. 102 acres. 11. If a teacher is now 57 years old, and has taught 38 years, at what age did he begin to teach ? Ans. 19 years. 12. How old was a person in 1865 who was bom in 1789 ? Ans. 76 years. 13. If I had $625 in a bank, and withdrew $249, Avhat re- mained? Ans. $376 4^. Illustrative Example, III. From 20000 take 9. Opeuation. (I) (9) (9) (9) (10) Here we have no tens to reduce to units, no hun- 2 dreds, and no thousands. We must then take one of the 9 2 ten-thousands (leaving 1 ten-thousand), and reduce • it to thousands, making 10 thousands. Reducing one of the thousands to hundreds, one of the hundreds to tens, and one of the tens to units, we leave 9 thousands, 9 hundreds, 9 lens, and have 10 units, from which, if we take 9 units, 1 unit will re- main. Having no tens to take from 9 tens, no hundreds to take from 9 hundreds, no thousands to take from 9 thousands, and no ten-thou- sands to take from 1 ten-thousand, we write these figures in their respective places below the line, and have for a remainder 19991. 43. Examples. 1. From 2017 years take 1028 years. Ans, 989 years. 2. A man, who had 1205 yards of cloth, sold 429 yards. How many yards were left? Ans. 776 yards. 3. There are 205 sheep in a flock ; if 109 of them should be driven to market, how many would remain ? A?is. 96 sheep. 4. A merchant bought goods for $1084, and sold them for $177 less than he gave ; how much did he receive for them ? Ans. $907. 28 SIMPLE NUMBERS. 5. 30070 men went into battle ; 4564 were slain, and 1300 were taken prisoners ; how many were left ? Ans. 24,206 men. 6. Take 229 oxen from 2006 oxen. Ans. 1777 oxen. 7. Subtract 25 hundred from 81 thousand. A7is. 78,500. 8. How many more in 47000 than in 702 ? Ans. 46,298. 9. 47000 less 46298 equals how many ? Ans. 702.. 10. 9832147 less 3472108 equals how many? 11. What number added to 9213628 will give 23475310 ? 12. What number subtracted from 7654321 will leave 369 ? 13. 86293210 minus 329876 equals how many ? 14. 987621085 — 329875232 =z how many ? 15. Find the sum of the last five answers. A7is. 771,984,860. i 6. 360080 + 7002 — 72824 = what ? 17. 3478921 + 368754 — 2878796 = what ? 18. From 7654321 — 1234567 take 53899. 19. From4673214 + 2792 take 98264. 20. 98432231 — 32636841 — 808994 = what ? 21. 8087670 — 7549094 — 89699 =z what ? 22. Find the sum of the last six answers. Ans. 77,642,007. 23. What is the difference between 19360742 and 9643278? 24. How many times can I take 7642 gallons from 21002 gal- lons, and what will remain ? 25. If the minuend is 36 quadrillion and the subtrahend 95 million 86, what is the remainder. Ans. 35,999,999,904,999,914. 26. If the minuend be 69 trillion and the difference 85 bil- lion, what is the subtrahend? 27. Philadelphia was founded in 1682. In what year was ]S'ew York city settled, it having been settled 68 years before ? 28. Victoria ascended the throne of England in 1837. How many years has she reigned ? 29. Napoleon commenced his brilliant career in 1795. How many years before his final defeat in 1815? 30. The Israelites left Egypt in 1491 B. C.,and 40 years after entered the land of Canaan. In what year did that event happen ? 31. In the year 1851, London had 2362000 inhabitants; Pekin was estimated to have 1500000. How many more inhab- itants had London than Pekin ? SUBTRACTION. 29 32. The equatorial diameter of the earth is 41843330 feet, and the polar diameter 41704788 feet ; required the difference. 33. The population of St. Louis in 1850 was 77860, and in 1860, 160773 ; required the increase in 10 years. 34. James Nye has in his possession $172 ; he owes $28 to A, $SQ to B, and $19 to C. After paying his debts, what will remain ? 35. I have saved from my income $362, and have $2180 in government bonds ; how much more must I save that I may pur- chase a house worth $3500 ? 4:4:, General Review, No. 1. 1. Two persons, who are 200 miles apart, travel towards each other, one 46 miles, the other 51 miles a day ; how far apart will they be at the end of one day ? 2. If the above persons travel away from each other, how far apart will they be at the end of one day ? 3. A man gave to his eldest son $3575, to his youngest son $4680, and to his daughter $2495 less than to the youngest son ; his whole property was worth $20000 ; what sum remained ? 4. A ship, which was valued at $15590, was sold at a loss of $4975 ; what did she bring ? 5. If the subtrahend be 369 quadrillion, and the remainder 99 quadrillion 13 billion, what is the minuend ? 6. The difference between two numbers is 95478. The larger number is 148769 ; what is the smaller? 7. How many times can 18640 be subtracted from 46806, and what will remain ? 8. Which of the two numbers 15672 or 10560 is nearer to 13465, and. how much ? 9. From what number must 846 be taken twice to leave 15684? 10. To what number must 962 be added three times to make 8472? 11. Which is nearer to 348628, 63248 + 93264, or 600063 — 59321 ? O^ For Dictation Exercises, see Key. 0Q SIMPLE NUMBEllS. MULTIPLICATION. 4r5. Multiplication is the process of finding a number equal in value to one number taken as many times as there are units in another number. The number which is multiplied is called the Multiplicand, the number by which we multiply is called the Multiplier, and the result obtained is called the Product. The multiplicand and multiplier are often called factors of the product, from the Latin facio^ I make, because, being multi- plied together, they make up the product. The product is also said to be the multiple of the factors. Thus, 7 tinaes 6 m 42, Here, 7 is the multiplier, 6 the multiplicand, and 42 the product ; or 7 and 6 are the factors of 42, which is their multiple. The sign of multiplication is a small, oblique cross, X> read, timeSf or, multiplied hy. Thus, 7X6 may be read either 7 times 6, or 7 multiplied hy 6. In the former case 7 is the multiplier and 6 the multiplicand, while in the latter 6 is the multiplier and 7 the multiplicand. The product is the same, whichever is the multiplier. Note. — In the process of multiplication, the multiplier must be an nbstract number. We cannot multiply pencils by pencils, or pencils by apples, but either may be multiplied by an abstract number, and give a product of the same denomination as the concrete factor. (Art. 4.) 4:6. Illustrative Exampl*e, I. Multiply 2364 by 7. Seven times 4 units i=: 28 units r^ Operatiox. e oQtiA -MT ^r V a 2 tens and 8 units. We write the 8 „ , 5 2364 Multiplicand, ..,..,, p actors < - , , .y y in the units* place, and reserve th« ( 7 Multiplier. ^ « , , , 2 tens for the tens place. 7 times 6 fHultiple, 16548 Product. tens =: 42 tens, which, with the 2 reserved tens, = 44 tensr=:4 hun- dreds and 4 tens ; we write the 4 tens in the tens' place, and reserve the 4 hundreds for the hundreds' place. 7 times 3 hundreds =zz 21 MULTIPLICATION. 31 hundreds, which, with the 4 reserved hundreds, =: 25 hundreds .zi: 2 thousands -\- 5 hundreds ; we write the 5 hundreds in the hundreds' place, and reserve the 2 thousands for the thousands' place. 7 times 2 thousands m 14 thousands, which, with the 2 thousands reserved, i= 16 thousands = 1 ten-thousand -j- 6 thousands; we write the 6 thou- sands in the thousands' place, and the 1 ten-thousand in the ten-thou- sands' place, and thus obtain for our product 16548. Note. — This result might be obtained by finding the sum of the num- ber 2364: taken seven times; that is, by adding 2364 to itself six times. Hence, Multiplication may be regarded as a short way of performing Addition. 47. Examples. Midtipli/ 1. 267 by 2 ; by 3 ; by 4 ; and add the products. Ans. 2,403. 2. 628 by 5 ; by 6 ; by 7 ; " " " Ans, 11,304. 3. 3401 by 8; by 9; " " " Ans, 57,817. 4. 90021 by 10; by 11; " " " ^^^s. 1,890,441. 5. 66285by 12; by8; " " " Jbz*. 1,325,700. 6. 4364 X 8 r= what.? 7. 7762 X 9 = what? 8. 5391 X 4 = what? 9. 3409 X 5 = what? 10. 9832 X 7 = what? 11. 8349 X 6 = what? 12. 22078 X 11 = what? 13. 19869 X 12 == what? 14. Add the last eight products, and multiply by 7. Ans, 5,205,081. 15. 123456 X 6 = ? 16. 987654321 X 7 — ? 17. Add the last two ;jroducts- Ans, 6,914,320,983 48. Illustrative Example, II. Multiply 3648 by 294. Operatiox. Here we are to multiply, not only by units, but 3648 ty tens and hundreds. We write the numbers 294 units under units, tens under tens, &c., and mul- -lA^Q^ tiply first by the units, as before, and then by the 32832 ^^^^ ^* ^^ evident that the product of any num- 7296 ^^^ multiplied by tens will be ten times as great as if multiplied by the same number of units ; 1072512 Ans. multiplied by hundreds, one hundred times as great as if multiplied by units; multiplied hv 32 SIMPLE NUMBERS. thousands, one thousand times as great, etc. Hence, when a number is multiplied by tens, hundreds, or thousands, the products thus ob- tained are written one, two, or three places farther to the left than when multiplied by units ; or, in other words, we multiply by the other terms as we multiply by the units, placing the fii'st figure of each product under the term by which we multiply. The sum of these partial products is the entire product. Hence the Rule for Multiplication. Write the multiplier under the multiplicand. Beginning at the right, multiply each term of the multiplicand hj each term of the multiplier, successively, placing the right hand Jigure of each partial product under the term hy which you multiply, car- rying as in addition. Add all the partial products, and the result will he the entire product. 4:^. Proof I. Take the multiplicand for the multiplier, and the multiplier for tlie midtiplicand. If the residt thus obtained be like the first result, the work is probably correct, 50. Proof II. By casting out the 9's. Thia method is much the easier, though not always sure. Note. — To cast out the 9's from any number, commence at the hft^ and add the digits towards the right. When their sum equals 9 or more, reject 9 and add the remainder to the next digit, and so on. The last remainder is called the excess of 9's. To Prove Multiplication by casting out the 9*s. Cast out the d's from each of the factors. U^ien multiply the remainders, shoidd there be any, cast out the d^sfram the product, and note the last remainder. Cast out the d^sfrom the answer, and if the remainder equals the one obtained above, the work may be presumed to oe right ; thus, 36184 3 + 6 = 0. 1 -f 8 =: 0. 4, 1st remainder. 2681 2 + 6 + 8 = 16z=04-7. 7-|-l = 8, 2d remainder. 36184 32 289472 3 + 2 zzz 5, last remainder. 217104 72368 97009304 Ans. 7 -{-. 3 ir:: 10 = + 1. 1+4 = 5, whicla equal- ling the remainder above, the work is right. Note. — For demonstration of rule, see Appendix. MULTIPLICATION. 33 Examples. 51, Perform and prove the following examples: — 1. 3G84X 3G = ? Ans. 132,G24. 2. 28-12 X 28 rir ? .^ Ans. 79,576. 3. 18762 X 236=r? 4. 128124 X 402=z:? 5. 189003 X 836==? 6. 12053 X 972 = ? 7. Add the answers to the last four examples, and multiply the sum by 3798. Ans. 857,040,363,792. 8. Multiply 123456789 by 98765. «>2, Any number may be multiplied by 10, 100, 1000, or a unit of any order, hi/ annexing as many zeros to the multiplicand as there are zeros in the multiplier, and j)lacin(/ the decimal point at the right. Examples. 9. Multiply 68432 by 10, by 100, 10000, 1000, 1000000, and add the products. , Ans. 69,192,279,520. 10. Multiply 3682 by 10000, 10, 1000, 100, 100000, and add the products. 03. Illustrative Example, III. Multiply 68432 by 86000. OrERATIOX. 68432 Here, by multiplying first by 86, and then annex- 86000 jj^g three zeros, which multiplies the first product b^ 410592 one thousand, the true result is obtained, and labor 547456 saved. 5885152000 Ans. Illustrative Example, IV. Multiply 832000 by 210. Opkration. 832000 Here the zeros in both the multiplicand and multi- 210 plier are disregarded until after muliplying the other gg2 terms together. 1664 174720000 Ans. 3 34 SIMPLE NUMBERS. 54i, Examples. 11. 6320X80 = ? 12. 4682 X 360 = ? 13. 92873 X 86300=? 14. 76000 X 8020 = ? 15. 32680 X 900100 = ? 16. 9876002 X 10001 = ? 17. 32001 X 20206 = ? 18. 987987 X 654653 = ? 19. 368043 X 77665=? 20. 23698 X 84293 = ? 21. Add the last ten answers, and multiply the sum by 100. Ans. 81,482,871,584,800. 22. How many hills of corn have I in my cornfield, which con- tains 97 rows and 45 hills in a row ? 23. If each hill produces 18 ears, how many ears does the field produce ? 24. I have four corn bins, containing severally 63 bushels, 54 bushels, 37 bushels, and 29 bushels. There are four pecks in a bushel. How many pecks do they all hold ? 25. Allowing 23 ears of corn to a peck, how many ears are there in the bins ? 26. If a barrel of flour costs 9 dollars, what will 368 barrels cost? 27. If a person by working 12 hours a day can do a piece of work in 37 days, in how many days can he do it working 1 hour a day ? 28. I have 5 bins, which contain 69 bushels each. What will be the capacity of a bin which will contain as much as all of them ? 29. If 6 yards of cloth will make one pair of shirts, how many yards will make one dozen or 12 shirts? How many will make 8 dozen ? 30. What will 3 dozen cost at 15 cents per yard for the cloth, 30 cents apiece for bosoms, wristbands, and buttons, and 50 cents apiece for making ? 31. It takes 7 yards of ticking for a single bed-sack; what must I pay for cloth for 18 single bed-sacks, at 16 cents per yard ? 32. If sheeting can be bought for 17 cents a yard, what must I pay for cloth for 21 sheets, allowing 10 yards for a pair? DIVISION. 35 33. "What will be the cost of 9 dressing gowns at 5 dollars apiece, 3 pairs slippers at 1 dollar a pair, 2 pairs boots at 4 dol- lars a pair, and 3 dozen stockings at 2 dollars a dozen ? 34. Suppose in 1 yard of cloth there are 580 fibres of warp ^ind 432 of filling, and that* each fibre of warp contains 32 strands, and each of filling 48, how many strands in the yard ? 35. The Lawrence Pacific Mills turn out material for about 65000 dresses in a week ; how many will they make in a year^ or 52 weeks? 36. Allowing 12 yards to a dress, how many yards do they make in a year ? E^ For Contractions in Multiplication, see Appendix. ^^ For Dictation Exercises, see Key. DIVISION. S5m Division is the process of ascertaining how many times ^ne number is contained in another, or of finding one of the equal parts of a number. Note. — In the example, " John has 10 apples, which he wishes to give to as many boys as he can, giving them 2 apples apiece, to how many can he give them ? " — it is evident he can give them to as many boys as 2 ts contained times in 10. In the example, ♦'!£ 16 pears are divided equally among 4 boys, how many pears does 1 boy receive ? " it is evident that 1 boy must receive one fourth of v/hat the 4 boys receive, or one fourth of 16 pears ; that is, one of tliefour equal parts of the number, 16 pears. The number which is divided is called the Dividend, the num • ber by which we divide is called the Divisor, and the result the Quotient, from the Latin quoties, how many times. The sign of Division is a short horizontal line between two dots, -^ ; thus, 9-^3 shows that 9 is to be divided by 3. Some- times the dividend and divisor take the place of the dots; thus, f. This expression may be read, 9 divided by 3, nine thirds, or one third of nine, and is the fractional * form of division. * See Art. 82. 35 SIMPLE NUMBERS. Short Division. Note. — This method is to be preferred where the divisor is not greater than 12. 06. Illustrative Example, I. Divide 936 by 6. Operation. We place the divisor at the left of the Divisor 6) 93 G Dividend, dividend, from uhich we separate it by .a — 7- curved line, and, drawing a straight line Quotient 156 beneath the dividend, proceed thus : 6 is contained in 9 hundreds 1 hundred times, with 3 hundreds remaining. We write the 1 hundred beneath the hundreds in the dividend, and reduce the 3 hundreds remaining to tens. 3 hundreds equal 30 tens, which, with the 3 tens of the dividend, equal 33 tens. 6 in 33 tens, 5 tens times, with a remainder of 3 tens ; writing the 3 tens in the tens' place, and reducing the remainder as before, we have 36 units. 6 in 36, 6 times ; writing the 6 in the units' place, we have 156 as the quotient of 936 divided by 6. Illustrative Example, II. Divide 17869 by 7. Opekatiox. In this example, as 7 is not con trained in 7 ) 17869 1 (ten thousand) any number of (ten thou- . '~Z77Z~ ^ T.^_ .,^^ sand) times, we shall have no ten thousands Ans. 2oa2-5Ilemamder. , / i ^u p . -, .^ m the quotient, and therefore take 17 (thousands) for our first partial dividend. We find also that the div- idend does not contain the divisor an exact number of times, but that there is a remainder of 5. As this does not contain 7 any whole number of times, we can indicate the division by placing the 5 in the quotient above the divisor, and have for the answer 25524, which is read, two thousand five hundred fifty-two and five sevenths. From the above examples we derive the Rule for SfioiiT Division. Beginning at the left, divide the •first term or terms of the dividend hy the divisor, make the residt the first term of the quotient. Prefix the remainder, should there he any, to the next term of the dividend, divide as before, and thus cojitinue till all the terms of the dividend are divided. Should there be a remainder after the last division, place the divisor beneath it, and annex the remit to the quotient. DIVISION. 37 57, Proof I. Division is the converse of Multiplication, the divisor and quotient being factors of the dividend : hence, to prove an example in division, multiply the quotient hy the divisor^ and to the product add the remainder. The sum thus obtained should equal the dividend. «58. Examples. Divide 1. 3G945 by 3. Ans. 12,315. 2. 987654 by 4. ^725.246,9131. 3. 864024 by 6. 4. 369801 by 9. 5. 120087 by 11. 6. 906102 by 3. 7. Find the sum of the last four answers. Ans. 498,044. 8. Divide 10101019 by 7. 9. Divide 16444006488 by 4. 10. 23456983241 -^ 9 = ? 11. 30089043921 -4- 7 n= ? 786491 12. = what? 8 369472 13. = what ? o 14. How many barrels of flour, at 7 dollars a barrel, can I buy for 259 dollars ? 15. At 11 cents a yard, how many yards of cloth can I buy for 368972 cents ? 16. If 12 pieces of cloth contain 408 yards, how many yards in a piece ? 17. Plow many weeks are there in 4781 days ? 18. How many hours will it take me to walk lo78 miles, at 5 miles an hour ? 19. 9 times a certain number equals 324783 ; what is that number? Ans. 36,087. 20. 8 X what = 36924? 21. 12 X what — 46817 ? Long Division. •59, Long Division is the process of dividing where the divisor is large, and the work written down. 60. Illustrative Example, L Divide 85232 by 23. 38 SIMPLE NUMBERS. Opeuatiox. 23 is contained in 85 (thous.) 3 (thous.) times ; 23')85*'32(3705ii. ^^ place the 3 (thous.) in the quotient, at the 69 " right of the dividend. 3 (thous.) X 23 =: 69 ~~ (thous.), which, subtracted from 85 (thous.), leaves a remainder of 16 (thous.). Bringing down the next figure of the dividend, we have 162 (hund's), which contains 23, 7 (hund's) times ; we place the ___ 7 (hund's) in the quotient at the right of the 3 17 (thous.). 7 (hund's) X 23 = 161 (hund's), which, subtracted from 162 (hund's), leaves 1 (hund.). Bringing down the 3 (tens) of the dividend, we have 13 (tens), which does not contain 23 any number of (tens) times. Placing a zero, therefore, in the ten's place in the quotient, we bring down the next figure, 2, and have 132 units ; 23 in 132, 5 times. "Writing the 5 in the unit's place in the quotient, multiplying and subtracting as before, we have a remainder of 17, and for our answer, 3705^|. Hence the Rule for Long Division. Beginning at the left, divide the first term or terms of the dividend hy the divisor ; make the result the first term of the quotient. Multiply the divisor hy this term, and subtract the product from that part of the dividend used. Annex the next term of the dividend to the remainder ; divide as before, and thus continue till all the terms of the dividend are divided. Should there he a remainder after the last division, place the divisor beneath it, and cmnex the result to the quotient. Note 1. "WTien it is difficult to determine the quotient figure at sight, trial divisors may be used. For example, divide 29847 by 476. .„^ V r.r..>. , It is evident that 476 is contained in the dividend 476 ^ 29847 ( • ^ ^ fewer times than 400 is contained in it, and more times than 500. Rejecting two right hand figures from the divisor, also from that part of the divi4end first considered, we see that 4 is contained in 29, 7 times, and 5 in 29, 5 times ; therefore the quotient figure cannot be more than 7 nor less than 5. Note 2. If, at any time, the product obtained by multiplying the divisor by any term of the quotient, exceeds the partial dividend, the quotient figure is too large. If, at any time, the remainder equals or exceeds the divisor, the quotient figure is too small. DIVISION. Bd 61. Proof II. By casting out the 9's. (Art. 5[), note.) Multiply the excess of 9's in the divisor, by the excess of 9's in the quotient, and find the excess of 9's in the product ; if it equals the excess of 9's in the dividend after the remainder has been subtracted, the work is presumed to be right. 62. Illustrative Example, II. Divide 26874 by 44. Operation. Proof. 44)26874 (610 Quotient. 4-f 4 = 8 264 6+ 1 = 7 47 7 X 8 = 06. 5 + 6 = 11=^:0 44 + 2. Casting out the 9, we not3 the remainder 2. 34 Remainder. 26874 — 34 zz: 26840. Placing the remainder, 34, in 2 + 6 + 8=16=0+7. a fractional form, in the quo- 7 + 4 = 11=0 + 2. This last ro. tient, we have for the answer. mainder, 2, being equal to the first, 610ff. the work is right. 63. Examples. 1. Divide 232848 by 56. Ans, 4158. 2. Divide 43572386 by 187. Arts. 233,007/gV. 3. Divide 18764321 by 262. 4. Divide 32456819 by 4618. 5. Divide 987654321 by 12345. 6. Divide 1459998 by 38, 19, 57, 171, 49, 513, 76842, and add the quotients. Ans. 182,075|f. 7. Divide 195989184 by 41, 16, 144, 164, 123,369, 72, 656, 1968, and add the quotients. Ans. 24,830,608. 8. Divide -43586118576 by 17, 56, 119, 136, 4158, 51, 72, 126, 99, 45738, 29106, 320166, and add the quotients. Ans. 6,288,215,934. 9. 975318642 +- 893 = ? 12. 8347300 +- 9004 = ? 10. 800000231 +- 73 = ? 13. 769800281 -+ 876 = ? 36845 , 3987659002 , , 11. -— ^ = what ? 14. — — -— — = what? 268 83297 40 SIMPLE NUMBERS. 04. When the divisor is 10, 100, or 1000, &c., we can divide by simply removing the decimal point in the dividend as many places towards the left as there are zeros in the divisor; the number at the right of the point will he the remainder ; thus, 368.4- 100 = 3.68 (or ^-f^^). Hence, if the divisor consists of any number of significant figures with zeros annexed, first cut off the zeros from the divisor and an equal number of figures from the right of the dividend, then divide what remains of the dividend by what remains of the divisor. To the remainder, if any, annex the figures that were cut off in the dividend ; thus, 38|00)968|42(2o ^. y ,. , i -, . ., ' -ft Disregarding the tens and units, we find now many times 38 (hund.) is contained in 968 208 (hund.), which is 25, with a remainder of 18 190 (hund.) ; this, with the 4 tens and 2 units left in the dividend, makes the entire remainder 1842, 1842 ... 3800 is contained in 96842, 25^A1 times. 60, Examples. 15. Divide 42179 by 1000; by 18000. Ans. 42xWo ; ^fi^\% 16. Divide 76532102 by 48O0; by 91000. 17. Divide 98000269 by 32600000; by 980000. 18. Bought a farm for $18715, at ^95 an acre; how many pcres were there in the farm? Ans. 197. 19. Paid $4505 for 27 acres of woodland; what was the price per acre ? Solution. If 27 acres cost $4505, one acre will cost one twenty- seventh of $4505, which equals, &c. 20. Paid $35328 for 368 acres of land; find the price per acre. 21. The distance from Boston to Albany is 198 miles, from Albany to Buffalo, 298 miles. How long will it take a train to pass over the road at the rate of 28 miles an hour, allowing 2 hours for detentions between Boston and Albany, 1 hour at Al- bany, and 3 between Albany and Buffalo ? Ans. 23|^ hours. 22. The Ohio Canal descends 1832 feet in 152 locks ; what is the average descent in each lock ? DIVISION. 41 23. If 8 presses can coin 19000 pieces of money in an hour, how many pieces can one press coin in a minute, 60 minutes making an hour ? 24. In how many days, of 12 hours each, can the president of a bank sign 9000000 bank notes, if he signs 8 in a minute ? ^^^For Contractions in Division, see Appendix. 1^ For Dictation Exercises, see Key. I 66, Questions for Review. 1. What is Arithmetic ? What are numbers ? What is an ab- stract number ? a concrete ? What is a unit ? Define Notation and Numeration. How are numbers represented ? Describe the Roman method; — the Arabic. Which is more used? Why is this sometimes called the Decimal System ? What is the decimal point ? By what is the number of units of any order expressed? By what is the order of units expressed ? 2. How do you write numbers ? How do you read numbers ? Name the first seven periods. Name others as far as you can. How are these periods separated ? What are the names of the places of each period ? 3. What is the least number of figures that will express units? — thousands ? — billions ? — trillions ? — millions ? — quadrillions. 4. In 189654238761, what is the largest number of thousands? — of millions ? — of ten-millions ? — of hundred-billions ? — of trillions ? — of tens ? — of hundreds — of ten-thousands ? 5. How will zeros at the right of a number affect it ? — at the left ? 6. What does a figure in the first place at the right of the decimal point represent? — in the second place? — in the third? 7. What is Addition*? What is the sign for Addition ? — for Equality ? How do you arrange numbers to be added ? Is this ab- solutely necessary ? Perform and explain an example containing four numbers of at least seven figures each. Give the rule. 8. AVhat is Subtraction? Name and define the terms used. What is the sign for Subtraction? Take 3684 from 7000068, and explain. Give the rule ; — the proof. When the minuend and differ- ence are given, how can you find the subtrahend ? When the subtra- hend and difference are given, how can you find the minuend? 9. What is Multiplication? Name and define the terms used. What is the sign for Multiplication ? Perform and explain an example in which the multiplier has, at least, two figures. Give the rule •. — . first method of proof j — second method. How do you multiply by i2 SIMPLE NUMBERS. 10, 100, 1000, &c. ? How do you proceed if there are zeros at the Vight of the multiplicand or multiplier? Tens X units z=: what ? Thousands X tens ? Hundreds X tens ? Ten-thousands X hundreds? Ten-thousands X ten-thousands? 10. What is Division ? Name and define the terms used. "What \s the sign for Division ? Perform and explain an example in short division. Give the rule. Give the proof by multiplication. Perforn? and explain an example in long division. Give the rule. Give the proof by casting out the 9's. 11. How do you divide by 10, 100, 1000, &c. ? How do you di- vide when the divisor contains zeros at the right of significant figures ? When the dividend and quotient are given, how can you find the divisor? When the divisor and quotient are given, how can you find the dividend ? When the multiplier and product are given, how can you find the multiplicand. When the multiplicand and product are given, how can you find the multiplier ? 67. Miscellaneous Examples. 1. Add nine billion, six hundred ninety-two million, eighty-one thousand sixty-four ; eighty-nine trillion, six hundred thirty-two million, ninety-one thousand eighteen ; eighty-seven thousand thirty-four ; and two hundred sixty-eight quadrillion, nine hun^ dred eighty-four trillion, ninety-eight million, one thousand ninety-four. 2. From (900362840218 — 986234681) take (7682 + 9619875.) * 3. Multiply (3684291 + 3642) by (8643264 — 8321628.) 4. Divide (3687291 — 86) by (3684 + 232.) 5. If 892 is one factor, and 28544 the product, \vhat is the other factor ? 6. 365 times what number z= 298570? 7. If the dividend is 38493, and the divisor 4277, what is the quotient ? 8. If the dividend = 42777, and the quotient 9, what is the divisor ? 9. There were 52 schools in Antigua in 1858, with 4467 scholars ; required the average number in each. * In examples 2, 3, and 4, first perform the operations indicated within the parentheses. MISCELLANEOUS. 43 10. David was born 1085 years B. C, and Washington 1732 A. D. ; what time elapsed between these events ? 11. What do I save a year, my income being $1600 a year, and my expenses $24 a week, 52 weeks making the year ? 12. Illinois produced in 1860, 1515594 pounds of maple sugar ; what was its value at 8 cents per pound ? 13. Mississippi produced 1195699 bales of cotton ; what was its value at 13 cents per pound, 400 pounds to the bale ? 14. Missouri produced 4164 tons of lead, worth $356660 > what was the value per ton ? 15. The population of Chicago in 1850 was 29963 ; in 1860, 109260 ; what was the average increase a year ? 16. If 8 men can do a piece of work in 24 days, how long will it take one man to do it ? 17. If 768 be one factor, and 861—237 the other factor, what is the product ? 18. Smith & Co. consume 74 tons of coal in a year ; how much more must they pay for their coal in 1864, when coal is $14 a ton, than in 1860, when it was $8 a ton? 19. From the invention of parchment to the invention of paper was 782 years ; to the use of quills in writing 741 years more ; to the invention of printing, 804 years more ; to the in- vention of stereotyping, 345 years more ; how many years from the invention of parchment to that of stereotyping ? 20. Parchment was invented 887 years B. C. ; when was paper invented ? Ans, 105 B. C. 21. When were quills first used in writing ? Ans. A. D. 636. 22. When was printing invented ? 23. When was stereotyping invented ? 24. 76854 divided by what number, gives a quotient of 56 and a remainder of 22 ? 25. What number divided by 87, gives a quotient of 3842 and a remainder of 76 ? 26. In 1853, Wheeler & Wilson made 799 sewing machines; in 1854, 956; in 1855, 1171; in 1856, 2210; in 1857, 45^1; in 1858, 7978 ; in 1859, 21306 ; in 1860, 19265 ; in 1861, 19725. Required the amount. 44 SIMPLE NUMBERS. 27. If a sewing machine can set 640 stitches in a minute, how many can it set in an hour? — in a day of 12 hours? — in 6 working days, or a week ? — in 52 weeks, or a year ? 28t There Avas sent to the U. S. mint, from 1823 to 1836, $4377984 worth of gold ; what was the average value sent a year? If gold was worth 16 dollars an ounce, how many pounds were sent, allowing 12 ounces to a pound ? 29* In the Pacific Mills, 200000000 gallons of water are used in a day. How many weeks would it take a man to pump it if he could pump a gallon in six strokes of the pump, 20 strokes a minute for 16 hours a day, allowing 6 working days per week? 30t If the earth is 95000000 of miles from the sun, and the moon at its full is 224000 miles farther on, and light travels at the rate of 191500 miles a second, how many seconds is it in passing from the sun to the moon and back to the earth ? Ans, AdSj%\^^^% seconds. 31. If 3871 be divided by 79, and the quotient be multiplied by 133, to this product 6523 be added, the amount divided by 40, and that quotient multiplied by 970, what will be the product? A?is. 316,220. 32. (17 — 2)^3=:?t 33. (7 + 3) X 2 = ?t 34. (1803 4-7982) X 3 t_^ 35. 19360 -^ 9^+1 + 43t 7 ""' 368 ~' 36.* (2 + 1 X 7 + 4) ^ 5 + (8 + 6) X 2 = ?t 37r (81 + 9) -^10 + 67 + (2 + 3X 7 + 7) -^6t=I? 1^^ For Dictation Exercises, see Key. t A vinculum, , or parenthesis (^ ), signifies that the same oper- Jition is to be performed upon all the quantities thus connected. In solving examples, it is generally better first to reduce all quantities con- nected by a vinculum, or parenthesis, to their simplest forms. Thus, in Ex. 32. (17— 2) -f- 3 =15-^3=5. Ex. 33. (7+3) X 2 = 10 X 2 = 20. Ex. 36. (2+ 1 x7 + 4)~54-(8 4-6)X2 = (3 X 7 H- 4) -i- 6 -i-; 14 X 2 = 5 -1- 28 = 33. Note. — Examples with stars are " optional examples." They may be omitted by younger pupils until a review, or altogether, if the teacher prefers. FEDERAL MONEY. 45 FEDERAL MONEY. 68, Federal Money is the medium of exchange in the United States. Federal is derived from the Latin fcedus, a league ; the money being used by states leagued or united under one government. Federal money consists of eagles, rep*, resented by E. ; dollars, represented by $ ; dimes, by d. ; cent&j. by cts., and mills by m. 2hUe of United States Currency, or Federal Money, 10 m. = 1 ct. 10 cts. r= 1 d. 10 d. = $1 $10 = 1 E. 60, As these denominate numbers increase and decrease like simple numbers, hy a scale of tens, they are written as sim- ple numbers are written, and operations are performed upon them as upon simple numbers, the dollar being regarded as the unit. The sign for dollar, $ , is placed before any number which we wish to designate as representing United States currency. s S s i „• §= =3 .§ g 2 W Q « o S $89,445 In business operations the denominations eagles and dimes are commonly disregarded, eagles being considered tens of dollars, and dimes, tens of cents ; thus, the above illustration is read 8d dollars, 44 cents, 5 mills. Examples. 70. Write the following: — 1. Seven hundred sixty-four dollars eighteen cents four mills. Ans, #764184. 2. 972 dollars 17 cents 2 mills. 3. 5768 dollars 9 cents 2 mills. 4. 10 thousand dollars sixty cents. 5. 9 million dollars 9 mills. Read the following: — <^2789.842. 11. $2009147.00. $9872.406. 12. $98765481.052.- $9084.007. 13. $4897.007. $864201.90. 14. $987801.94. 8329871.045. 15. $81746.807. 46 FEDERAL MONEY. 71 6. 7. 8. 9. 10. 16. What is the largest number of cents contained in exam- ple 6? — 7? — 8? — 9?— 10? IstAns. 278,984 cts. 17. What is the largest number of dimes ? — of mills ? — of eagles? 1*^ Ans. 27,898 dimes. 18. Read examples 11 to 15, making cents the unit of numer- ation. 19. Reduce $86452. to cents ; to mills. Ans. 8,645,200 cents ; 86,452,000 mills. 20. Reduce $9841.72 to mills. 21. Reduce 8712647 cents to dollars. 22. Reduce 3687514 mills to dollars. How do you reduce dollars to cents ? to mills ? How do you reduce mills to dollars ? to cents ? How do you reduce cents to mills ? 23. $9843.621 + $4687.32 + $84,321 + $.07 -f- $.64 -f $973,241 — ? A?is. $15,589,213. Note. — In addition and subtraction of Federal Money, dollars should be written under. dollars, cents under cents, etc. 24. $3684.271 + $765.42 + $1763.417 + $8645.217 — 3.68 = ? . Ans. $14,854,645. 25. From $8643.271 + $98367.489 take ($37,862 + $33695.41). 26. From $3471.009 — $.71 take ($987,541 + $862.73). 27. From $4645. + $8178. take ($9827. — $6712.86). 28. $34865.002 X 46 = ? Ans. $1,603,790,092. Note. — In the example above, as mills are multiplied, the answer must be mills. 29. 11 X $3687.40 = ? 31. $98417.83 X 791 = ? SO. $946,918 X 478 zr:? 32. 984 X $7654216.69=? FEDERAL MONEY. 47 Note . — It will be obvious that in the four following examples, tht quo- tient must be of the same denomination as the dividend. 33. $13428. -^ 9 = ? 35. $241364.48 -^- 56 = ? 34. $7352.88 -^ 12 = ? 36. $3712471.712 -i-ASS=z? ^^^ Illustrative Example. $1725. -M8=? Opekatiox. 18) 1725 ( 95.833^V Ajis. 162 "TTt In this example, after dividing the dollars, we have f.^ a remainder of 15 dollars; this we reduce to dimes — by annexing a zero, and dividing, obtain 8 dimes for the quotient figure, and have a remainder of 6 }^^ . dimes, which we reduce to cents and divide, and have 60 a remainder of 6 cents, which we reduce to mills and 64 divide, and have a remainder of 6 mills, and for the (50 entire quotient, $95,833^%. Ans. 54 6 •Note. — In the four following examples continue the division to mills, 37. Divide $9867. by 37 ; by 91 ; by 416. 38. Divide $89000. by 17 ; by 42 ; by 368. 39. Divide $36421.90 by 18 ; by 48. 40. Divide $6003489. by 96 ; by 543. 41. How many times are $.34 contained in $36.72? 42. How many times are $.25 contained in $645. ? Note. — In dividing Federal Money by Federal Money, when the de- nominations are unlike, it is necessary first to reduce the dividend and divisor to the same denomination. The answer will be an abstract num- ber ; thus, $645. -J- $.25 =r 64500 -^ 25 = 2580. 43. Divide $186432.18 by $0,032. 44. Divide $382971.21 by $93. 45. Bought 1 pair of boots for $1.37; 1 pair for $1.65.; slippers for $.95 ; shoes for $.65 ; and shoes for $.82. Required the entire cost, j^ r)\l\ ^| 46. Bought a horse for $95.00 ; a wagon for $63.00, and harness for $ 15.00 ; kept them a week, paying $ 2.50 for board 48 FEDERAL MONEY. for the horse, then sold them all for $ 175.00. Did I gain or lose, and how much ? ,S \Q c Tj 47. What cost 8 pairs geese at $1.28 per pair? j^v| i^^'^ 48. Bought 2 dozen pigeons at $.85 per dozen, 2 dozen at $1.10 per dozen, and 1 dozen for $.90. What should I pay? ^/7i 49. 8874 sheep were sold at $4.13 per head; what did they bring ? 50. There were shipped to Great Britain in one year from New York, 20602243 pounds of butter. What would it bring at 15 cents per pound ? 51. 39479897 pounds of cheese were shipped the same year. Kequired the receipts at 7 cents per pound ? 52. 4778 beeves were sold in New York market in one week, averaging 874 lbs. apiece, at 7 cents per pound ; what was received for them ? 53. Bought 2 pieces of flannel, each containing 62 yards, for $39.68, and sold them for 40 cents per yard. What did I gain ? 54. Paid a man $16.25 for 13 days' work; what was that a day? bb. Paid $5.10 for 17 boxes strawberries; what was that a box? 56. Among how many boys may $10 be distributed, that each may receive $0,625 ? 57. Sold 35 barrels Greenings at $1.75 per barrel, 17 barrels Baldwins at $1.80 per barrel, 12 barrels fall Harveys at $1.25 per barrel, and 25 of Russets at $2.25 per barrel. Paid 17 cents a barrel for picking, and $12.00 for transportation. What re- mained after all my bills were paid ? 58. Paid $3.00 for 1 dozen apple trees, $3.36 for 1 dozen peach trees, $3.30 for one half dozen pear trees ; what did I pay for the whole, and how much a piece for each kind ? 59. Paid a carpenter for stock and work for a house, $450.75 ; for mason's work, $38.25 ; for digging and stoning cellar, $47.18 ; for painting, $40.00 ; to the plumber, $8,125. I then sold it, and lost, in so doing, $14,305 ; what did I sell it for ? Ans. $570. FEDERAL MONEY. 49 60. Bought a farm, containing 40 acres meadow and 17 wood- land, for $2850.00. Sold to one man 10 acres woodland for $85.00 per acre ; to another a house lot of one acre for $90.00 ; and the remainder to a third for $2025.00. What did I gain by the operation; and for how much per acre did I sell the re- mainder? Ans. $11^ i $44.02^85. Bills. 73. When, in a business transaction, one person receives money, property, or services from another, he becomes indebted or is debtor for the amount he receives. The person who parts with the money, property, or services, is credited for the amount he has given, and hence is called the creditor. A written statement of the amount of the debt, with the items included, is called a hill, and is usually written in forms like those on the following pages. When the creditor is paid the amount due, he acknowledges the receipt by his signature at the foot of the bill, after the words " Received payment." A bill thus signed is said to be receipted, 74 • Find the cost of each article in the following bills, and their several amounts. Mr. James Crocker 10 bbls. St. Louis Flour, Buffalo, November 10, 1862. > Bought of Henry Shedd, extra, at $ 9.50 12 " Western « medium, " 7.75 8 " Canada " extra, " 6.72 14 « Canada « choice extra, *' 7.87 3 « Corn Meal, it 4.25 20 bu. Northern Oats, a .61 Received payment, $376.89. Henry Shedd, By George Ba^^ 50 FEDERAL MONEY. (2.) Lawrence, November 18, 1862. Mr. D. Danforth, 65 bu. Potatoes, at Bought of J. Smith, $0.55 300 lbs. Squashes, " 450 « Pork, « .01 .11 35 bu. Beans, " 2.50 85 « Rye flour, " 2.25 Received payment, $367.00 Ans. J. Smith. (3.) Boston, April 17, 1863. Mr. James Blake, Bought of Breck & Co., 3 bu. Herds Grass, at $2.25 75 lbs. Clover Seed, " .11 25 bu. Canary Seed, " 18 lbs. Mustard Seed, « 3.62 .13 25 « Hops, « 22 " Hops, " .17 .16 Received payment. $115.61 Ans. Joseph Breck, for Breck & Co. (4.) New Bedford, October 9, 1862. Mr. J. L. Rice, To Henry' Brown, JDr.* To 2 bbls. Pork, prime, at $ 15.50 « 250 lbs. Hams, « .09 « 475 " Butter, « .24 « 4^2 " « « .18 Received payment, $ Henry Brown. means that Mr. Rice is debtor to Mr. Brown. Dr, is rea4 BILLS. 51 (5.) New York, , March 7, 1868. A. M. Phipps, Esq., To Samuel Sloane, Dr, Jan. 7. To 12 lbs. Tartaric Acid, at $ .85 ii ii ii 7 " Blue Vitriol, " .25 « 12. a 3 oz. Morphine, " 7.00 " 13. ii 5 " Quinine, " 4.00 Feb. 2. ii 2 lbs. Cardamoms, " 3.50 a a it 10 « Cream Tartar, " .51 ii a a 8 « Cubebs, .53 a a a 5 " Gum Copal, " .68 ii ii a 8galls.Cod Liver Oil, " 1.75 Received payment, $ Samuel Sloane. (6.)* Bristol, January 1, 1863. Otis Butler, Esq., 1862. To Ralph Burnside, Dr. Apr. 3. To 96 lbs. Rice, at $ .07 a a " 3 « Saleratus, li .08 a a " 28 « Castile Soap, ii .16 May 9. " 25 « Pearl Starch, a .09 « 15. " 196 « Crushed Sugar, a .13 a a « 196 « Brown Havana, ii .12 June 5. « 46 " Hyson Tea, a 1.12 July 10. " 37 " Gunpowder Tea, a .95 $ Cr.t May 10. By 1 Wagon, $42.00 " 16: " 2 Cows, at $35.00, ti a " Cash, 10.00 $ $. Received payment, •♦r • Ralph Burnside. t This means that Mr. Butler is credited for goods or cash delivered. Cr. is read " creditor." 52 FEDERAL MONEY. 75. Find the amounts due in the following examples, and make out the bills. 7. Charles Fuller purchased of James Munroe, Jan. 4, 1863, 1 horse for $95.00, 2 cows at $50 apiece, 1 wagon for $62.00, 2 shovels at $1.12 apiece, 30 bushels corn at $.65 per bushel, and 17 bushels wheat at $1.62 per bushel. 8. Samuel Banks sold to Abraham Seward, March 10, 1863, 2 pieces flannel, of 62 yards each, at $.49 per yard ; 5 pieces cotton, bleached, at $.24 per yard, 2 of the pieces containing 36 yards each, and 3 containing 35 yards each; 38 yards ticking, at $.29 ; 86 yards brown sheeting, at $.27 ; 42 yards broadcloth, at $3.65 per yard. 9. Dr. Cardamom bought of James Mortar 3 gallons castor-oil at $2.50 ; 9 pounds oil peppermint at $2.50 ; 4 pounds oil cassia at $3.62 ; 4 pounds oil orange at $3 ; 6 pounds oil lemon at $4.25 ; 5 pounds oxalic acid at $.33 ; and 5 pounds Seneca root at $.95- 10. Baldwin & Lewis, of Cincinnati, bought of Balch & Ray- ner, Boston, 24 sack coats at $15.75 ; 36 vests at $3.50 ; 95 pairs pants at $4.38 ; 4 dozen pairs suspenders at 42 cts. per pair ; 23 dozen pairs gloves at 68 cts. per pair; and 15 dozen collars at 13 cts. apiece. 11. Hiram Teachwell bought of Mark Thrifty, Nov. 8, 1862, 2 Dictionaries, at $.87 apiece; 9 Yocal Cultures, at $.70; 12 Walton's First Steps, at $.13, and 24 Worcester's Spellers, at $.20. Dec. 2, he bought 2 reams paper at $2.12, 3 dozen pen- cils at $.50, and 12 slates at $.17. Dec. 10, he paid Mr. Thrifty $20.00, and Jan. 1, 1863, Mr. Thrifty made .out his bill. Re- quired the balance due. 12f Solomon Katchall bought of Hiram Southack, Aug. 11, 1862, 12 pairs congress gaiters, at $2.75 ; 12 pairs misses' gaiters, at $1.12; 8 pairs kip boots, at $2.75; 12 pairs, at $.95; 9 pairs boys' metallic-toed shoes, at $.72; 12 pairs gents.' boots, at $6.75; 12 pairs, at $4.25; 5 pairs, at $3.15. He sold Mr. Southack 18 yards black silk, at $1.17; 48 yards brown sheeting, at $.19; 18 yards crash, at $.13, and 20 yards flannel, at $.45. J^^ For Dictation Exercises, see Key. ANALYSIS. ^i ANALYSIS. 7G, Analysis in arithmetic consists in determining the solu- tiosn of an example from the relations of the numbers given in that example. The given number which is. of the same denomination as the required answer forms the basis of all the reasoning, and should be the first written in perlbrming an example. The value of any number of things may be obtained by first finding the value of a single thing or unit of the same denom- ination. This unit is sometimes called the unit of computation. Illustrative Example. If 25 barrels of flour cost $175, what cost 17 barrels ? Operatiox. $175 is the term of the same denom- l'<^5 I7::=i^ll9 Ans ^^^^^^^ ^^ ^^® required answer. Before iJ5 finding the value of 17 barrels, we must know the value of 1 barrel.* If 25 barrels cost $175, 1 barrel will cost 1 twenty-fifth of $175, and 17 barrels will cost 17 X 1 twenty-fifth of $175,=: $119. Examples. 1. If 13 acres of land produce 780 bushels of com, how many bushels will 5 acres produce ? -^ns. 300, 2. If 5 boxes of oranges cost $21.80, what cost 21 boxes ? Ans. $91.56. 3. If a car runs 207 miles in 9 hours, how far will it run in 25 hours ? 4. If 18 rows of potatoes yield 54 bushels, how many bushels will 405 similar rows yield ? 5. If $19.74 were paid for 14 bushels of rye, what must be paid for 25 bushels ? 6. If 19 tons of coal run an engine 266 miles, how far will 14 tons run it? 7. If 5 oxen consume 85 pounds of hay in 1 day, how much will be required for 1 yoke of oxen of the same size, and iox the game time ? * 1 barrel is the unit of computation. 54 ANALYSIS AND REVIEW. 8. How many pounds of coffee can be bought for $15, if 40 lbs. cost $8 ? Note. — If $8 pay for 40 pounds, $1 will pay for 1 eighth of 40 pounds, and $15 will pay for 15 X 1 eighth of 40 pounds = 75 pounds. 9. If 150 barrels of apples were bought for $200 and sold for $350, what would be gained by selling 45 barrels at the same rate ? 10. If a quantity of hay lasts 22 oxen 105 days, how many days will it last 5 yoke ? Note. — If it lasts 22 oxen 105 days, it will last 1 yoke 11 X 105, and it will last 5 yoke 1 fifth of 11 X 105 days =231 days. 11. A field of wheat was reaped by 10 men in 6 days ; what length of time would be required for 15 men to reap the same amount ? 12. A cistern can be emptied in 35 minutes by 7 pipes ; in what time can it be emptied, if 5 only of the pipes are open ? 13. If 1423 operatives can do a piece of work in 12 days, in what time will 2400 operatives perform the same work ? 14. If a certain piece of work can be performed by 250 men in 14 weeks, how many more must be employed to perform it in a week ? 15. A garrison of 10000 men have provision to last them 6 weeks ; if 2000 men be killed in a sally, how long will the provisions last the remainder? 77. Questions for Review. 1. Federal Money. What are the denominations of federal money? Give the table. How do you write numbers in federal currency ? What is considered the unit ? Give the sign for dollars. How do you reduce eagles to dollars ? dollars to cents ? dollars to mills ? cents to mills ? mills to dollars ? 2. How do you add numbers in this currency ? How do you sub- tract ? When you multiply, of what denomination is the product ? When you divide by an abstract number, of what denomination is the quotient? Divide $185 by 7, continue the division to mills, and explain. What is necessary in order to divide mills by dollars ? by cents ? In dividing cents by dollars, is the quotient abstract or con- REVIEW. 55 cT3te? In dividing dollars by an abstract number, is the quotient abstract or concrete ? 3. Bills. What is a bill? Ans. It is a writing given by the cred- itor to the debtor, showing the amount of the debt. Who is the creditor ? the debtor ? What is the receipt of a bill ? 4. Analysis. What is analysis ? ^^^'hich number forms the basis of the reasoning. T8, General Review, No. 2. 1. 287 4- 5 million -f- 36 thousand -f- 59481 z= ? 2. Add 567 to the sum of the following numbers: 121 ; 232; 343; 454; 565; 676; 787; 898. 3. Take 987 from each of the following numbers, and add the re- mainders : 9876 ; 6678 ; 3644 ; 7573 ; 2432 ; 4001. 4. What number must be added to the difference between 58 and 7003 to equal 938425 ? 5. What number, taken from the quotient of 1833000-^47 leaves 25 ? 6. What number equals the product of 1785, 394, and (624—48) ? 7. If 5872 is the multiplicand, and half that number the multiplier, what is the product? 8. If 4832796 is the product, and 1208199 the multiplicand, what is the multiplier ? 9. If 894869 is the minuend, and the sum of all the numbers in the third example is the subtrahend, what is the remainder ? 10. If 700150 is the dividend, and 3685 the quotient, what is the divisor ? 11. If 28936 is the divisor, and 86 is the quotient, what is the dividend ? 12. Divide 87 million by 15 thousand. 13. $3.75 4- $9.32 + $.75 + $10. +$2.185 -}- 4 cents =:? 14. $19.— $.75— $8.25 + $3.54zzz? 15. From 18 X $5,873, take $3.68 + 4. 16. If $183.30 is the dividend, and $3.90 the divisor, what is the quotient ? 17. If $98 60 is the dividend, and 17 the divisor, what is the quotient ? J^^ For changes, see Key. 56 PROPERTIES OF NUMBERS. PROPERTIES OF NUMBERS. 79. Signs. — Recapitulation. -f- signifies plus, or more. rr signifies equal to. — signifies minus, or less. X signifies multiplied by. ^ signifies greater than. -~- signifies divided hy, <^ signifies less than. .•. signifies therefore. () parenthesis, and , vinculum, signify that the same op-, eration is to be performed upon all the quantities thus connected. Definitions. 80. Numbers are either integral or fractional. 81. Integral numbers, or Integers, are whole numbers. 8S. iPractional numbers are parts of whole numbers. 83. A Factor or Divisor of a number is any number which is contained in it without a remainder ; thus, 2 is a fac- tor of 6. 84. A Prime Number is a number which contains no integral factor but itself and 1 ; as, 1, 2, 3, 11. 85. A Composite Number is a number which contains other integral factors besides itself and 1 ; as, 4, 6, 8, 25. 86. A Prime Factor is a factor which is a prime number. 8T. A composite number equals the product of all its prime factors ; thus, 12 = 2X2X3. 88. Two numbers are said to be prime to each other when they contain no common factor except 1 ; thus, 8 and 15 are prime to each other. 80, The Power of a number is the number itself, or the prod- uct obtained by taking that number a number of times as a factor. The number itself is the first power; if it is taken twice as a factor, the product is called the second power^ or square; if three times, it is called the third power ^ or cuhe ; if four times, the fourth power, &c. Thus, the second power of 3 is 3 X ^ = 9 ; the third power of 3 is 3x3X3 = 27; the fiflh power of 3 is 3X3X3X3X3 = 243. DIVISIBILITY OF NUMBERS. 57 90, Tlie Index or Exponent of a power is a figure which shows how many times the number is taken as a factor. It is written at the right of the number, and above the line. Thus, in 5^, 7^, 2'*, the exponent ^ shows that 5 is taken three times as a factor, ~ that 7 is taken twice, and 4 that 2 is taken four times as a factor. ©1, The Root of a number is one of the equal factors which produce that number. If it is one of the two equal factors, it is the second, or square root ; if one of the three, the third, or cube root; if one of the four, the fourth root, &c. Thus the square root of 9 is 3, the cube root of 125 is 5. 9^. V ^s the Radical Sign, and, by itself, denotes the square root ; with a figure placed above, it denotes the root of that degree indicated by the figure ; thus, ^ signifies the third root, -^ the sixth root. Divisibility of Numbeks. 93, (1.) Any number whose unit figure is 0, 2, 4, 6, or 8, is even. (2.) Any number whose unit figure is 1, 3, 5, 7, or 9, is odd. (3.) Any even number is divisible by 2. (4.) Any number is divisible by 3 when the sum of its digits is divisible by 3 ; thus, 2814 is divisible by 3, for 2 + 8 + 1 + 4 p=. 15, is divisible by 3. (5.) Any number is divisible by 4, when its tens and units are divisible by 4 ; for, as 1 hundred, and consequently any num- ber of hundreds, is divisible by 4, the divisibility of the given number by 4 must depend upon the tens and units ; thus, 86324 is divisible by 4, while 6831 is not. (G.) Any number is divisible by 5 if the units* figure is either S or ; for, as 1 ten, and consequently any number of tens, is divisible by 5, the divisibility of the given number by 5 must depend upon the units. (7.) Any number is divisible by 6, if divisible by 3 and by 2. (8.) Any number is divisible by 8, if its hundreds, tens, and enits are divisible by 8 ; for, as 1 thousand, and consequently any 58 I'HOPERTIES OF NUilBEIiS. number of tliousands is divisible by 8, tlie divisibility of the given number by 8 must depend on the hundreds, tens, and units. (9.) Any number is divisible by 9 if the sum of its digits is divisible by 9*; thus, 368451 is divisible by 9, and 23476 is not. (10.) Any number is divisible by 10, 100, or 1000, if it con- tain at the right 1, 2, or 3 zeros ; and so on. (11.) Any number is divisible by 11, if the difference between the sums of the alternate digits is 0, or a number divisible by 11 ; thus, in 126896, as (1 + 6 + 9) — (2 -f-8 + 6) =: 0, the number is divisible by 11 ; and in 9053, as (9 + 5) — (0 + 3) z= 11, the number is divisible by 11. (12.) A number is divisible by any composite number, if it is divisible by all the factors of that number. 03, There are no rules of sufficient practical importance for determining when numbers are divisible by other numbers than those spoken of above. Their divisibility must be ascertained by trial. To do this, — Divide the number successively hy higher and higher primes, until one is found which divides it, or until the quotient is smcdler than the divisor. If no divisor is then found, the number is prime; for, if a number contain any prime factor greater than its square root, its corresponding factor must be less. 94. If the odd numbers are written in order, and every third one from 3, every fifth one from 5, every seventh one from 7, and so on, be marked, and the figures 3, 5, 7, &c., be written under the figures as they are marked, the remaining numbers will be primes, and those marked will have for their factors the numbers written beneath ; f thus, — 1, 3, 5, 7, 0, 11, 13, X$, 17, 19, M, 23, ^^, ^t, 29, 31, 3 3,5 3,7 5 3,9 0;gl, ^$, 37, 210, 41, 43,^^, 47, ^0, $X, &c. 3,11 6,7 3,13 3,5 7 3,17 9,16 * See Appendix. f Eratosthenes, in the third century B. C, discovered this method of finding primes and factors of numbers, and as he made his table of parch- ment, cutting out the composite numbers as he found them, this parchment wai called Eratosthenes' Hiece. TABLES OF PRIME AND COMPOSITE NUMBERS. 59 By applying this principle, a tabic can easily be made of the primes and of the composites, with their factors. Table of Pkimk Nu.mbkiis to 1201. I (51 151 251 359 403 593 701 827 953 1069 2 GZ 157 257 3G7^ 4G7 599 709 829 967 1087 3 71 1()3 2(53 373 479 GOl 719 839 971 1091 5 73 1G7 2(>9 379 487 G07 727 853 977 1093 7 79 173 271 383 491 G13 733 857 983 1097 11 83 179 277 389 499 G17 739 859 991 1103 13 89 181 281 397 603 G19 743 863 997 1109 17 97 191 283 401 509 G31 751 877 1009 1117 19 101 193 293 409 621 G41 757 881 1013 1123 23 103 197 307 419 523 643 761 883 1019 1129 29 107 199 311 421 541 647 769 887 1021 1151 31 109 211 313 431 547 G53 773 907 1031 1153 37 113 223 317 433 657 659 787 911 1033 1163 41 127 227 331 439 5G3 661 797 919 1039 1171 43 131 229 337 443 5G9 673 809 929 1049 1181 47 137 233 347 449 671 677 811 937 1051 1187 53 139 239 349 457 577 683 821 941 1061 1193 59 149 241 353 461 587 691 823 947 1063 1201 Table of the Composite Numbers to 917, Which contain no prime factor less than 7 {excepting 1*). Nos. Factors. Nos. Factors. Nos. Factors. Nos. Factors. Nos. Factors. 49 72 289 17« 469 7, 67 623 7,89 779 19,41 77 7, 11 299 13, 23 473 11, 43 629 17, 37 781 11,71 91 7, 13 301 7, 43 481 13, 37 637 7', 13 791 7, 113 il9 7, 17 319 11, 29 493 17, 29 649 11, 59 793 13, 61 121 112 323 17, 19 497 7, 71 667 23, 29. 799 17,47 133 7, 19 329 7, 47 511 7, 73 671 11, 61 803 11, 73 143 11, 13 341 11, 31 517 11, 47 679 7, 97 817 19, 43 161 7, 23 343 7^ 527 17, 31 689 13, 53 833 72, 17 169 13« 361 192 529 23'<» 697 17, 41 841 292 187 11, 17 371 7,53 533 13, 41 703 19, 37 847 7, 11» 203 7, 29 377 13, 29 539 7^ 11 707 7, 101 851 23, 37 209 l.l, 19 391 17, 23 551 19, 29 713 23, 31 869 11, 79 217 7, 31 403 13, 31 553 7, 79 721 7, 103 889 7, 127 221 13, 17 407 11, 37 559 13, 43 731 17, 43 893 19, 47 247 13, 19 413 7, 59 581 7, 83 737 11, 67 899 29, 31 253 11, 23 427 7, 61 583 11,53 749 7, 107 901 17, 53 259 7,37 437 19, 23 589 19, 31 763 7, 109 913 11, 83 287 7, 41 451 11, 41 611 13, 47 767 13, 59 917 7, 131 * 1 is a factor of all nTimbers. 60 PROrERTIES OF NUMBERS. FACTORING OF NUMBERS. Oo. Illustrative Example, I. Resclve 48 into its prime factors. OrERATION. 48 =6X8; 6 = 2X3; 8 = 2X2X2;.-. 48 = 2X 2 X 2 X 2 X 3, or 2^* X 3. Hence, Rule I. To resolve a number into its prime factors. — • First separate it into any tivo factors ; separate these factors, if they are composite, into others, and so on, till all are prime. Proof. Multiply the factors thus obtained together, and the product, if the work is correct, will equal the given number. 96. Examples. Resolve the following numbers into their prime factors. — 1. 32. Ans. 25. 2. 84. Ans. 22 X 3 X 7. 3. 88. Ans. 23 X 11. 4. 56. 7. 100. 5. 49. 8. 150. 6. 72. 9 69. 10. 11. 12. 81. 99. 144. 13. 14. 15. 64. 77. 108. 16. 130 17. 125. 18. 250a 07o Illustrative Example, II. Resolve 42075 into its prime factors. Operation. 3 ) 42075 3 ) 14025 5)4675 Here we divide, successively, by such prime numbers as will leave no remainder, till we obtain a prime number for a quotient ; since the product, of these prime numbers, 3, 3, 5, 5^ 11, and 17 equals the given number, they must be the prime factors of that number. Hence, 5)935 11)187 17 iw5. 32 52 11, 17. Rule IL Divide th. number by any prime number which is ^ntained in it without a remainder. Divide the quotient in the same manner, and thus continue till a quotient is obtained which is a prime number. This quotient and the several divisors are the ^rime factor:-. GllEATEST COMMON DIVISOR. 61 Note. — Tlie work may sometimes be shortened by dividing by a com- posite number, remembering afterwards to substitute the factors of that number for the number itself. Thus, in the above we may divide by 9 instead of dividing by 3 twice. 98, Examples. Resolve the following numbers into their prime factors. 19. 17G. A71S. 2^ X 11. 20. 180. Ans. 2^ X 3^ X 5. 21. 192. Ans. 2« X 3. 22. 208. Ans. 2^ X 13. 00, Select the prime numbers in the columns below, and find the factors of the composite numbers. 23. 260. 27. 357. 24. 285. 28. 644. 25. 329. 29. 684. 26. 338. 30. 2310. 1. 341. 6. 450. 11. 704. 16. 947. 2. 344. 7. 590. 12. 719. 17. 971. 3. 362. 8. 560. 13. 769. 18. 2681. 4. 367. 9. 596. 14. 808. 19. 1163. 5. 409. 10. 689. 15. 839. 20. 3248. J^* Por Dictation Exercises, see Key. Greatest Common Divisor. 100. A L-ommon Divisor of two or more numbers is any number that will exactly divide each of them ; thus, ^ is a common divisor of 12 and 18. 101. The Greatest Coinmoii Divisor is the greatest num- ber that will exactly divide each of them ; thus, 6 is the great- est common divisor of 12 and 18. 103, Illustrative Example. Find the greatest common divisor of 12, 30, and 42. Operation. -^ oyoyq As 2 and 3 are the only common fac- QQ 2 X 3 X ■'* ^^^^ ^^ "^^' ^^' ^^^ ^^' ^^ follows that ^2 2X3x7* 2 X 3, or 6, is the greatest common di- G C. D. ^ 2 X 3 zir ^Ans. ^'^''^' H^^^^' Rule I. To find the greatest common divisor of two or (52 GREATEST COMMON DIVISOR. more numbers : Separate the numhers into their prime facton^ and jind the product of such as are common. J 103. Examples. Find the G. C. D.* of 1. 48, 56, and 60. Atis. 4. 2. 24, 42, and 54. Ans. 6. 3. 108, 45, 18, and 63. 4. 18,36, 12, 48, and 42 Note. — In Example 4, 18 is a factor of 36, and 12 of 48. The G.C. D. of 18 and 12 must be the G. C. D. of 18, 12, and their multiples, 36 and 48 ; .-. we need only find the G. C. D. of 18, 12, and 42. Find the G. C. D. of 5. 42, 28, and 84. 6. 26, 52, and 65. 7. 32, 18, 108, and 25. 8. 114, 102, 78, and 66. 104:. When numbers cannot readily be separated into their factors, the following method may be adopted : — Illustrative Example. Find the G. C. D. of 91 and 325. Operation. We divide 325 by 91, to see if it is a 9A ) 325 ( 3 divisor of 325, for 91 is the greatest divisor 273 of itself; if it is a divisor of 325, it is the 52 ) 91 ( 1 G. C. D. of 91 and 325. It is not a divisor 52 of 325, for there is a remainder of 52. 52 39 ^ 62(\ ^^ ^^^ greatest divisor of itself; if it is a 39 divisor of 91, it is the G. C. B. of 52 and ~V\\ oq / q ^1' It is not a divisor of 91, for there is a OQ remainder of 39 ; 39 is the greatest divisor of itself; if it is a divisor of 52, it is the ^^ G. C. D. of 39 and 52. It is not a divisor of 52, for there is a remainder of 13 ; 13 is the G. C. D. of itself and 39. It must therefore be of 39 and 52, for 52 := 1 X 39 + 13. If it is the G. C. D. of 39 and 52, it must be of 52 and 91, for 91 z=l X 52 + 39. If it is the G. C. D. of 52 and 91, it must be of 91 and 325, for 325 = 3 X 91 -f 52. Hence the following : Rule IT. To find the G. C. D. of two numbers: Divide the greater number hy the less, and the less number by the re- mainder, if there is any, and thus proceed, dividing the last * Greatest Common Divisor, GREATEST COMMON DIVISOR. G3 livisor hy the last remainder, until nothing remains. The last iivisor is the G. O. I), sought. To find the G. C. D. of more than two numbers, Jind the G. C. D. of ariy two of them, and then of that divisor and a third number, and so on till all the numhej's are taken. 10«5« Examples. Find the G. C. D. of 12. 229 and 954. 13. 392, 1008, and 224. 14. 6581, 6611, and 249. 9. 198 and 297. Ans. 99. 10. 222 and 564. Aiis. 6. 11. 529, 782, and 1127. Ans. 23. 15. What is the width of the widest carpeting that will ex- actly fit either of two halls, 45 feet and 33 feet wide respect- ively ? Ans. 3 ft. 16. A has a piece of ground 90 feet long and 42 feet wide; what is the length of the longest rails that will exactly suit its length and its width ? Ans. 6 ft. 17. A lady has one flower bed measuring 10 feet around, and another measuring 18 feet. If she borders the beds with pinks, what is the greatest distance she can set her pink roots apart, and have them equally distant in the two beds ? Ans. 2 ft. 18. A man has 90 bushels Kidney potatoes, 60 bushels Jack- son Whites, and 105 bushels Red Rileys. If he puts them all into the largest bins of equal size that will exactly measure either lot, how many bushels will each of his bins contain ? 19. What is the length of the longest stepping-stones that will exactly fit 3 streets, 72, 51, and 87 feet wide, respectively ? 20. What is tlve length of the longest curb-stones that will exactly fit 4 strips of sidewalk, the first being 273 feet long, the *^cond 294, the third 567, and the fourth 651 ? 1^ For Dictation Exercises, see Key, A 64 COMMON FRACTIONS. FRACTIONS. 1^6* A Fraction is one or more of the equal parts of a unit; thus, f, read three fourths, shows that a unit has been divided into four equal parts, and that three of those parts are taken. lOT, The number which shows into how many equal parts \ unit is divided, is called the Denominator of the fraction, because it denominates or names the parts ; thus, 4 is the denom- inator of f . 108, The number which shows how many parts are taken, is called the Numerator ; thus, 3 is the numerator of f . 109, The numerator and denominator are called the Terms of a fraction. 110, A Common or Vulgar Fraction is a fraction whose denominator and numerator are both expressed, the numerator being written above, and the denominator below, a dividing line ; as, ^, |, /^. IfiS. A Decimal Fraction is one whose denominator is 10, 5 X 2^^^^ 10 r^ |"™|j 1 ' — I ^ ! — [ Fraction diminished. Proposition II. If we divide the nmnerator of a fraction bt a whole number, we divide the number of parts and thus diminish the value of the fraction ; but if we divide the denominator of a fraction, we divide the number which shows into how many parts the unit is divided, and thus increase the size of the parts, and consequently increase the value of the fraction. ^ I I M I H 2-^2 1 Illustration, "l — 6 |— ( — j 1 — | — } — | Fraction diminisx^ed. 2 2 g _^ 2 ~— - 3 h"n^ 1 ' Fraction increased. Proposition III. If we multiply the numerator and denom- inator, each by the same number, we increase the number of parts of the fraction, but diminish their size in the same propor- tion ; consequently the value of the fraction is not altered. Q ILLUSTRATION * Fraction not altered ?J A^is. many units as | is contained times in ^g^, which is 4|- times. Hence the Rule. To reduce an improper fraction to a mixed number; — Divide the numerator hy the denominator. MULTIPLICATION OF FRACTIONS. 71 Examples. Change to whole or mixed numbers, 9_8 "5 • Alls. 4:^. Ans. 211. Ans. 19^. 6. 7. 8. 9. 10. 865 ■Er2T- 3 7J)_4 8, 11. 12. 13. 14. 15. J96 F8- 1^" For Dictation Exercises, see Key. A 138. Multiplication of Fractions by Whole Numbers. As multiplying the numerator of a fraction multiplies the number of parts, their size remaining the same, and dividing the denominator multiplies the size of the parts, their number' re- maining the same (Art. 119), it follows that, — To multiply a fraction by a w^hole number, we may either multiply the numerator by the whole number, or divide the denom- inator. The latter method is preferable when the denominator can be divided without a remainder, as it gives the answer in lower terms. III. Ex. Multiply | by 4. , 1st Operation. 7x4 28_„, . - — = - = 3|, Ans. 2D Operation. = ~z=z3h Ans. 8 °' 8-T-4 2 "We might have cancelled in the first operation, and thus have $ ' Multiply Examples. 2 1. i by 5. Atis. 2f. 9. ^\ by 19. 2. 3^ by 6. Ans. ^|. 10. t\\ by 21. 3. ^^ by 504. Ans. 23fi. 11. T%h by 95. 4. r\ by 15- 12. A^ by 54. 5. f by 4. 13. m by 274. 6. A by 110. 14. ■b\% by 328. 7. tVj by 9. 15. ^fk by 762. 8. U by 11. 16. _33^8^7JL by 55. 72 COMMON FRACTIONS. 17. If one yard of cloth co^ts | of a dollar, what will 17 yards cost ? 18. If a ton of coal costs | of an Eagle, how much will 15 tons cost? 19. Required the cost of 28 pounds of candles, at f of a dol- lar a pound. 20. Multiply 25G| by 18. Ans. 4623f. " 21. Multiply 376ff by 21. ^^ For Dictation Exercises, see Key. 1S9* Multiplication of Whole Numbers by Fractions. III. Ex. Multiply 8 by |. 4 OPERATION. 8 multipHed by 5 is 5 X 8 j if it is 0X 5 4X5 az J multiplied by |, a number one sixth as 3 ^' ' large as 5, the product must be one 3 sixth as large as if 5 had been the mul- tiplier, or i of 5 X 8. The expression then becomes 8 times 5, di^ided by 6; after cancelling,—^ z= — =: 6f , Ans. Hence the o o Rule. To multiply a whole number by a fraction ; — Multiply the whole number by the numerator of the fraction, and divide that product by the denominator. Examples. Multiply 1. 36 by |. Ans. 24. 2. 568 by |. Ans. 473^. 3. 385 by ^■^. 4. 3681 by ^V 5. 5432 by ^J. 6. 87036 by ^V 7. What cost I of 1 ton of hay, at $12 a ton-? Operation. If 1 ton of hay costs $12, ^ of a ton 1.50 will cost I of $ 12, and | of a ton will $ 3^^.00 X 5 __ ^ ^ ^^ j^^^^ cost 5 times | of $ 12. Cancelling, we $ • J • have $1.50X5 = $ 7.50, ^rw. 8. What cost f of an acre of land, at $100 an acre ? 9. What cost y^2 of an acre of land, at $150 an acre ? 10. What cost f of a gross of pens, at $.96 a gross ? 11. What cost 5^ cords of wood, at $7.56 a cord? 12. What cost 3| hogsheads of molasses, at $18.80 a hogshead ? MULTIPLICATION OF FRACTIONS. 73 13. What cost 2f firkins of butter, at $12.60 a firkin? 14. "What cost 63f yards of flannel, at $.54 a yard? 1^ For Dictation Exercises> see Key. 130* MULTIPLICATIOJf OP FRACTIONS BY FRACTIONS. III. Ex. Multiply | by |. Opehation, i multiplied by 8, is 8 times | ; if it be multi- 5 X ^ 20 plied by f , a number one ninth as large as 8, the ^ = rr? ^^* product must be one ninth as large as if 8 had 3 been the multiplier, or one ninth of 8 times |. The expression then becomes ^^ j after cancelling, g— -^ =. ^^ Ans. Hence the RcLfi, To multiply a fraction by a fraction ; — Multiply the numerators together for a new numerator^ cmd the denominators for a new denominator^ Examples. Mvdtiply 4. If by If. 5. If by l^f, 6. Ill by-VsV-- 7. _8^ X 21= what? Note. — Reduce mixed numbers to improper fractions before multiply- ing by fractions. 8. 18f X ^1 r= ? i 10. 15| X t\ =: ? 1. 1 by f- Ans. ^, 2. f by iV Ans. f J. 3. IS by ?• Ans, -^7f. 9. 5i2Xjf7- = ? I 11. 3|X31|=:r? Note, — In the ninth example, first reduce 1^ to lower terms. 12. 38§|X^f = ? J 13. 2^X111^? 14. What cost 2j^jy boxes of raisins, at If dollars a box ? 15. What cost 10 J tons of coal, at $7| a ton? 16. What cost bh pounds of coffee, at j^jj of a dollar a pound ? 17. What cost 11-^^ pounds of pork, at ^^ of a dollar a pound ? ^F" For Dictation Exercises, see Key. 131. Reduction of Compound Fractions to Simple Fractions. III. Ex., I. If 1 dollar buys f of a yard of cloth, how much will f of a dollar buy ? 74: COMMON FRACTIONS. Operation. Explanation 1st. If one dollar buys | 3X3 __ 9 J , of a yard of cloth, ^ of a dollar will buy ^ 6 X"4 "~ 20 5 •» * of I of a yard, and f of a dollar Avill buy 3 times ^ of I of a yard, zrz ^^^ of a yard. Explanation 2d. If one dollar buys | of a yard of cloth, | of a dollar will buy f of | of a yard. ^ of ^ of a yard is ^g- of a yard, which may be shown by dividing | of a yard into 4 equal parts. The whole yard can thus be divided into 4 X 5, or 20 equal parts ; there- fore, 1 part will be -^^ of the whole yard. If ^ of ^ := ^L, \ of f musf be -^^j and f of | must be 3 times ^3^, or -t^^. Ans. III. Ex., II. f of -f = what? L> JExercises, see Key, 84 COMMON FRACTIONS. 143( Eeduction of Fractions to Equivalent Feao TiONs having a Common Denominator. "When the denominators of fractions are alike, they are said to have a Common Denominator. III. Ex. Reduce |, |, and ^ to fractions having a common denominator. "We can change these fractions to fractions of any given denom- inator ; but tht most convenient denominator for most purposes is that which is the least common multiple of the denominators of tli% given fractions ; and, in the following examples, such denoraiflators are always required. In the preceding example, we must first find the L. C. M. of 3, 4, and 6, whjch is 12 ; and then reduce f , |, and | to twelfths. 1=1|, .-. |=r^ of If or -j^, and | = ii/^^z=-jV By the same process we fijid that f r= 3^, and | = \^. Ans. -^, -^-^^ i^. Hence the Rule. To reduce fractions to equivalent fractions having a common denominator : Reduce the fractions, to their simplest forms ; find the least common multiple of the denominators for the common denominator ; multiply the numerator of each fraction hy the number hy which you would multiply its denominator to produce the common denominator,"^ The respective products will he the numerators of the required fractions, III. Ex. Reduce ^, -j^l, and f to fractions having the L. C. D. Entire Opkration. 8 = 2X2X2 i^^xTEJ^ii. 12^:2X2X3 -i-V = ^ -W-^- - f f • 9=3X3 f = 2 x_2^x^2.x 2 — |6. L. C. M.=:2 X2X2X3X3=:72. Here 72 is the L. C. M. ; and as 8 = 2 X 2 X 2, it must be multi- plied by 3 X 3 to produce 72, .-. \ = J^^^, and | = .&i<^I = 4|. Show why yV = H ; why | z= ||. * If that number is not readily seen, it may be found by dividing the toramon denominator by the denominatpr of the priginal fraction. ADDITION OF FRACTIONS. 85 Examples. Reduce the following to fractions having the least common denominator: — 4. ^\, i§, and A-t 5- T5r ih 1^5-' and sV 6- /(J, tVi -3%^ and ^2-. 7 3 5 2 8 nnrl 3 2 1. J-, f , and f. 2. f Z^-. and /t- 3. t'ttj 2^, and -j^^. 1^=" For Dictation Exercises, see Key.' 143. Addition of Fractions Examples. X f + i^what? f-j-f zrivvhat? f + f =what? Ans, f. 4. 36^ -f ^ij- := what ? T^or + t5(J = what ? 6. A+ini — what? The above examples are easily performed, as the quantities to be operated upon are like quantities, that is, have the same de- nominator. In such cases, we have only to add the numerators. When fractions of different denominators are to be added, thii/ must first be reduced to fractions having a common denominator^ III. Ex. — Add J and ■^^. £ 16 £ 15 lO-i-15 n 9 36* 12 36* 36 36* Operation. ~z=z — —z=z—. Ans. 7. -1 + ^ + 1 = ? 9- f+A+i + l 10. f + l 4-^ + ^8=? 11. i*5 + A + i + l=? 12./. + l + i=? 13. A + /a + A = ? •ItV Am. 2^. 14- A + A + i!=? 15. « + A + tV = ? 16. f + S + 4§=? 17. A + a^ + A = ? 18. A+g\+A + ii=? 19*t'6 + A + 3'2+/T = ? Note. — Add the whole numbers and fractions of the following, and similar examples, separately. 21.36+21 + 7,^2=? 22. 18§ + 16^4-28f = ? 23. 272^+16^ + 181 = ? 24! 104^3^ + 8/0 + 480|i = ? t What operation should j&rst be performed upon these fractions ? 8(5 COMMON FRACTIONS. 25. 2007y3(y + 1070| + 8040| = ? 27.%vof2| + iof| + f of2^V = 28r2/^ + ^of/3 + ^ = ? 29MA + lT\+i|=what? Ans. 7^V% For Dictation Exercises, see Key. 144. Subtraction of Fractions. Examples. 1. f— 2-? ^,?«. f 5. TV^i-Tli = ? 2. i-| = ? Ans.^ = i. 6. t¥^-tIf = ? 3. j'^-j\ = ? 7.^-1 = ? 4. T^TT T§(T = ? Ans. ii. Note. — The denominators in Example 7 being unlike, the fractions must be reduced to fractions having the same denominator. 8. f-f = ? 9. ^- 10. A 11. 1-^^-? TT 3 "5 —■7 Ans. ^. Ans. ^V- T3- 12. A- -A — ? 13. A- -T^j: = ? 14. 2§- -li = = ? Note. fractions. Subtract without changing the mixed numbers to improper 18. 18^ — 15^3 = ? 19. 17^— 12^9^ = ?. ^715.43. 15. 8-L — 3f = ? Ans. 6^. 16. 7| — 2J^=r? Ans.b:^. 17. 103_5^|=:? Note. — As -^-^ cannot be taken from ■^, it will be necessary to reduce 1 of the 17 to Iialves, making the minuend 16|, when subtraction can be lasily performed. 20. 2^ — 1| = ? 2f Ans. ^. ? Ans. 14j-|. 21. 17i 22. 12^ — f r=? 23. 26f— If = ? 24. 19 — 2|z=? Ans. 16|. 25. 36 — 1=:? 26. 75 — lo^zzr? 27. l-^of^^ •g — ? 16 28. 18f-^f-i|of2^=i:what? For Dictation Exercises, see Key. ADDITION AND SUBTRACTION OF FRACTIONS. 87 14^. Addition and Subtraction of Fractions Combined. Give a rule for the addition of fractions ; for subtraction. Examples. '' 1. I + A-A = ? ^715 '■ m- 2. i- -1+1 = V 3. 1- -i + i- ■i+i- -i + i- i+i- -tV Ans. mi- 4. f- ■i+ii- -Ti^ = : what ? 5. 1- ■i-i- ■i-T\ — ^v-^v -tJ.: — ? 6. 1- ■i-A- "rJff — what? 7. 20- -S^ + t' rof§ = -p 8. 8/t -2| + ' H=? 9. Ul X f off + 10f— rVof M =? ^^,. 18^^.1.. 10. 7-(/^-^9^) = ? ^ J,z.. Gif. 11. 5-(t + 3V) = ? Ans.4iU. 12. A man receives 4^ per cent, commission for selling goods ; he pays | per cent, for storage ; what per cent, does lie retain ? 13. If he receives 6f per cent, for selling goods, and If per cent, for insuring their sale, and pays If per cent, for storage, and y^2- per cent, for auctioneering ; what per cent, does he retain ? 14. How much will be left of a piece of cloth containing 7 yards, after cutting from it 2 vests and a coat, allowing f of a yard for a vest and 4|^ yards for a coat ? 15. Bought of Mrs. Frye 1 bonnet for $4.37^, 2 hats at $2.12^ apiece, 4 yards ribbon at $.16f per yard, 2 yards ribbon at 33^ cents a yard, and gave in payment a ten dollar bill ; what should she give me in return ? 16. From 8 apple trees I gathered as follows: 2^ barrels, 5^ barrels, 5-| barrels, 4|- barrels, 3f barrels, If barrels, 3^ barrels, and 2^ barrels. I sold 15 J- barrels to one man, and 2^ barrels to another, how many barrels had I left ? 88 COMMON FRACTIONS. 17* To what must you add the difference between 8f and 36/^, that the amount may be 50§ ? 18f If 7| X f — '^Tua is the minuend, and ^^^ the remamder, what is the subtrahend ? 1416. Greatest Common Divisor of Fractions. * III. Ex. Find the greatest common divisor of f , |, and |. OPEPwVTION. G. C. D. of 6, 8, and 4 z= _2_ ^^^ - We find the G. C. D. of the L. C. M. of 7, 9, and 5 = S15' ' numerators 6, 8, and 4 to be 2. 2 is a divisor of 6, but must be divided by 7 to be a divisor of |. It must also be divided by 9 to be a divisor of f , and by 5 to be a divisor of ^. To be at the same time a divisor of these fractions, it must therefore be divided by 7, and 9, and 5, or by their least common ■^lultiple. Hence the Rule. To find the G. C. D. o^ f»*actions: Reduce the frac- tions to their lowest terms ; then divide the G. C. D, of the numer' cLtors by the L. C, M. of the denominators. EXAMPLI^.S. 1. Find the G. C. D. of §, |, and |. Ans. yV- 2. Find the G. C. D. of y^^, -r\, and 2^ or f . Ans. ^V 3. Find the G. C. D. of 3^, ^^, |, and J-f." 4. Find the G. C. D. of f , f , and 4. Note. — 4 can be regarded as i. 5. Find the G. C. D- of f J, ^3^, f, and 2. G. What is the size of the largest cup which is an exact meas- ure of 1^, 1|, 8^, and ^ pints ? 7. What is the width of the widest carpeting that will fit 4 rooms of the following widths: 13^ feet, 21 feet, 31^ iaeX, 3G2 feet ? ^^ For Dictation Exercises, see Key. 147. Least Common Multiple of Fractions.* III. Ex. Find the least common multiple of J-, f, and f . ^ __^ We find the"" L. C. M. of the Operation. L. C. M. of 1, 3, and 5 = 15 numerators, 1, 3, and 5, to be G. C. D. of 2, 4, and 6 z=.T' ^''^- ^^' ^^^ ^'^ ^^ "<^* ^^'^^^ *« ascertain the least number that * Articles 146 and 147 can be omitted by younger pupils. QUESTIONS FOR REVIEW. 89 will contain 1, 3, i nd 5, but one that will contain ^, |, and |. To cont iin each of these fractions separately, it might be divided by 2, by 4, or by 6 ; but to contain them at once, it can be divided only by theii- G. C. D. Hence the EuLE. To find the L. C. M. of fractions : Reduce the frac- tions to their lowest terms, then divide the L. C. M. of the numer- ators hy the G. C. D. of the denominators. Examples. 1. Find the L. C. M. of -^, JL3^ and 7^. Ans, 47 6f. 2. Find the L. C. M. of if, ^ of 3^, and 6. Ans. 390 3. What is the width of the narrowest cloth that can be cu* into strips either f , IJ^, or 4 inches wide ? 4. What will be the length of the shortest court that can b< paved with stones of either of the following lengths, viz., 1^ fl.r 2 ft., 4 ft., or 2§ ft. ? Ans, 24 it. 5. What must be the width of the narrowest court that will receive either of the same stones widthwise, their widths being 1 ft., 1^ ft., 3 fl., and 2 ft. ? 6. On a stringed instrument in perfect tune, while C makes 1 vibration, D makes f , E J, F |, G f , A ^, B VS and C 2. If all are struck at once, in how many vibrations of C will they all again coincide ? 7. In how many vibrations of C will C, E, G, and C coincide ? wUl C and D coincide ? C and E ? B and C ? C and O ? ^^ For Dictation Exercises, see Key. Questions for Review. Definitions and Properties of Numbers. — AVhat is the sign for plus ? for minus ? for greater than ? less than ? equal to ? multi- plied by? divided by? therefore? What does a parenthesis or vinculum signify ? What are integral numbers ? What are frac- tional numbers ? mixed numbers ? What is a prime number ? a com- posite number ? What are the factors of a number ? What is a prime factor ? A composite number equals what product ? When are num- bers prime to each other ? What is a power of a number ? What is the square or second power of a number ? the fifth power ? What is a root of a number ? the square root ? the cube root ? the sixth root ? 90 COMMON FRACTIONS. MTiat i s the sign for a power ? for a root ? What indicates the degree of root ? What is an even number ? an odd ? Divisibility of Numbers. — When are numbers divisible by 2 ? by 3P by 4P by 5P by 6? by 8? by 9? by 10? by 11? by any com- posite number ? How shall we ascertain whether any given number is prime ? Describe Eratosthenes' sieve ? Factoring of Numbers. — What is the simplest way of resolving numbers into their prime factors ? What other method can you de- scribe, and when would you use it ? Find the factors of 180 by first method, and explain the process. Find the factors of 10296 by sec- ond method, and explain the process. Greatest Common Divisor. — What is a divisor of a number ? a com. mon divisor of two or more numbers ? the greatest common divisor ? Find the G. C. D. of three numbers by the first method given. Explain and give the rule. Find the G. C. D. of three numbers by second method. Explain and give the rule. In what cases is the second method the better ? When is it necessary to find the G. C. D. of numbers ? Fractions. — What is a fraction ? Name and describe its terras. Name the different kinds of fractions of which you have learned. Define a common fraction ; a decimal fraction ; a proper fraction ; an improper fraction ; a mixed number ; a compound fraction ; a complej? fraction. Give an example of each. Explain the expression ^. Upon what does the value of a fraction depend ? Which of the fundamental rules is indicated by a fraction ? What effect does multiplying the numerator of a fraction have upon that fraction ? Why ? In what other way could you produce the same effect, and why ? What effect does dividing the numerator have upon a fraction ? Why ? In what other way could you produce the same effect, and why ? What effect does multiplying both terms of a fraction by the same ntimber have upon it ? AVhy ? What effect does dividing both terms of a fraction have upon it ? Why ? Reduction of Fractions. — How do you reduce fractions to lower terms ? What is cancellation ? How do you reduce whole or mixed numbers to improper fractions ? How do you reduce improper frac- tions to whole or mixed numbers ? Multiplication of Fractions. — How do you multiply a fraction by a whole number ? a mixed number by a whole number ? Explain, by an example, the method of multiplying a whole number by a fraction. Multiply a fraction by a fraction ; explain and give the rule. How do you multiply a mixed number by a mixed number or a fraction ? How MISCELLANEOUS EXAMPLES. 9X do you reduce compound fractions to simple ones ? Can you give one general rule for multiplying fractions, whole or mixed numbers, by frac- tions ? Division of Fractions. — How do you divide a fraction by a whole number ? a mixed number by a whole number ? a whole number by a fraction ? Explain, by an example, the method of dividing a fraction by a fraction, and give the rule. Give one general rule for dividing a fraction, a whole or mixed number by a fraction. How do you reduce complex fractions to simple ones ? How do you find what part of one number another is ? Least Common Multiple. — Define a multiple; a common multiple of two or more numbers ; the least common multiple. When do you make use of the L. C. M. ? Give and explain the first method of find- ing it; the second. What does the X. C. M. of prime numbers equal ? Common Denominator. — When are fractions said to have a com- mon denominator ? In what operations upon fractions do we first reduce them to those having the same denominator ? Can we change fractions to those of any denominator? How ? {Ans. By dividing or multiplying'the numerator by the same number by which we divide or multiply the denominator to produce the denominator required.) What denominator is generally chosen ? Reduce a simple, a compound, and a complex fraction to those of the same denominator, explain the process, and give the rule. Addition and Subtraction of Fractions.- — How do you add fractions of different denominators ? How do you subtract one frac- tion from another ? How do you add mixed numbers ? In subtraction of one mixed number from another, how do you proceed when the frac- tion in the subtrahend exceeds that in the minuend ? G. C. D. and L. C. M. of Fractions^ — How do you find the G. C. D. of fractions ? How do you find the L. C. M. of fractions ? Find the G. C. D. of I, f and {^, and explain. Find the L. C. M. of ^, |, and -^^, and explain. 14:8, Miscellaneous Examples. 1. Into strips of what widths may I cut cloth which is 36 inches wide, that none may be wasted, the width of the strips to be expressed in inches ? 2. How many gallons in the largest vessel which will exactly measure 3 hogsheads, containing severally 128, 94, and 158 gal- lons? * Optinal. 92 COMMON FRACTIONS. a What will 16^ yards of cloth cost at $.53 a yard ? 4 What cost 9^ bushels of corn, at $.87^ a bushel ? 5. What cost 27 If acres of land, at $31| per acre ? 6f f of I of 56 times what number equals ^§f ? 7. I paid $.65 for 2 boxes of strawberries ; what will be the cost of 45^ boxes at the same rate ? 8. What is my bill for 7 pear trees, $1 apiece for the trees, and $2 a dozen for setting ? 9. What do I receive per pound by selling 15f pounds of cof- fee for $3^? 10. -^ of a man's property is in land, and is valued at $2324| ; what is the value of his whole property ? 11. Bought f of an acre of land for $40.75 ; what would 1 acre cost at the same rate ? 12. What costs 3 pieces of calico, 37^ yards in a piece, at 19^ cents per yard ? 13. If 32y*2- acres of land cost $1100, what costs 1 acre? 14. Sold my house and farm of 47f acres for $6150 ; allow- ing $3500 for the house, what did I receive per acre for the* land? 15. How long will a barrel of flour last a family of 8 persons, if it lasts 3 persons 4^ months ? 16* What number is that from which if you take -^jj, the re- mainder will be ^V ? 17. What number is that to which if you add 9|, the sum will bel24|? 18. What is that number to which if you add f of 26^, the sum will be 147-^? 19. Bought 7^ yards broadcloth at $5 per yard, 14^ yards of kerseymere at $1;^ per yard, 4| yards of silk at $| per yard, and f yards of doeskin at $4^ per yard, for w^iich I gave in pay- ment a $100 bill. What balance is due me ? 20r I have paving stones 12 inches long and 10 inches wide; what must be the width of a walk which will just receive these stones, laid either lengthwise or widthwise ? 21* What is the smallest sum of money which can be exactly paid in pieces of money worth either $.16f or $.12^? MISCELLANEOUS EXAMPLES. 93 22. How long will 200 pounds of meat last 9 persons at the t-ate of 2| pounds a day for each person ? 23. What length of time would a man require to travel around the earth, the distance being 25000 miles, if he travel at the rate of 31^ miles per day? 24. If a man can build 2f rods of wall in a day, how much can he build in 6^ days ? 25. What is that number J of which exceeds ^ by 2 ? Note. ^ — | z= ^ ; if ^ be 2, |^ will equal 20 X 2 = 40, Ans, 26. What number is that f of which exceeds ^ by llf ? 27. How many bushels of wheat can a man purchase for $2724^, at 31f cents per bushel? 28. What is f -^ (yV of | of S^). 29. What is (§ of ^) -^ (^ of f ) ? 30. If I buy 125 bushels of corn at 41 § cents per bushel, and sell it at 521 cents per bushel, what do I gain? 31. AVhat number divided by | equals 125^? 32. What are the contents of 3 floors measuring as fol- lows : 13^ square yards, 32Y^g square yards, and 49|| square yards ? 33. The product of three numbers is 74^; two of them are 8f and 6^^^ ; what is the third ? 34. Exchanged 42 tubs of butter, averaging 48f pounds, at 21 J^ cents per pound, for 42 barrels of flour, at $9f per barrel, and received the balance in cash ; required the balance. 35. I have three boxes of cloth, each containing 12 pieces, each piece containing 4f yards, weighing 3^ pounds to the yard ; what is the weight of the whole ? . 36. What will 42| quires of paper weigh at f pound per quire? 37. A-grocer has five casks of raisins of the following weights : 115| pounds, 117f pounds, 9d^% pounds, 100^ pounds, and 121^ pounds ; what is the average weight per cask ? 38. What is the cost of the above at 8§ cents per pound ? 39. Owning f of a paper-mill, I sold f of my share for $1750 ; what is the value of the whole mill at the same rate ? 94 COMMON FRACTIONS. 40. A man sold 50 yards of cloth at the rate of Ij- yards for 2 dollars ; what did he receive for it ? 41t Mr. Gray raised 212 bushels of potatoes, f of which he stored in 4 equal-sized bins ; what did each bin hold ? He sold the other | at the rate of 3^ bushels for 2 dollars ; what did he receive for the potatoes which he sold ? 42. If -If of a ton of lead cost S134f, how many ounces of gold at SI 9^ per oz. will pay for one ton of lead ? 43. When hay was $15 per ton, I gave f of a ton for If tons of coal ; what was the coal worth per ton ? 44. If f of a yard of cloth will pay for six hats worth $4^ per dozen, what is the price of the cloth per yard ? 45. If /,y of an acre of land cost $280^, what will 5f acres cost? 46. If a man walks 9^ miles in 2^ hours, how far will he walk in 4^ hours ? 47. At the rate of 4^ miles an hour, what time Avill be re- quired to walk 122 miles? 48. In 1860, I purchased cotton at 8^ cents a pound, which I sold in 1862 at 90| cents. What did I gain on 1000 lbs. ? 49. If a man can earn $2^^^^ per day, how many days' work will he have to give for a suit of clothes, of which the coat cost $25^, the pants $8j\, and the vest $5^ ? 50f The longest canal in the world, is the Grand Canal in China; ^ -f ^ + ^V + ^i^ + tV of its length is 331§|| miles ; what is its entire length ? Ans. 650 miles. 51. If f of I of a ship cost $42,000, what is f of her worth? 52. In a certain manufactory, | of the operatives are Germans, I Dutch, j\j- Scotch, ^ English, -^^ Canadians, and the remainder, 140, native Americans ; what is the whole number, and the num- ber of each nationality ? 53. ^ of my money is in gold, ^ of the remainder in silver, and the balance, $360, is in bank notes ; how much money have I in all? 54t If 17 boxes of raspberries cost $2.83^, what part of a box can I buy for 12^ cents? ^^■r' MISCELLANEOUS EXAMPLES. 95 55. If a body in falling descends 16^'^ feet in the first second ()f time, three times 16jL in the next second, and five times 16^ teet in the third second, how far will it fall in three seconds ? 56. Owing a man in Paris 1325^ francs, I have shipped to him $375^ worth of rice. If the franc is worth 18f cents, how much liave I overpaid him in Federal money ? 57. 7f oz. of gold are to be divided among 3 men and a boy, — the boy to have half as much as a man ; what will each have ? 58.* If the wages of a man per month are $35^, and if the wages of 3 boys are equal to the wages of 2 men, what will be the wages, of 10 men and 30 boys for a month ? 59. What is that number to which if f of itself be added the sum will equal 64 ? Aris. 40. 60r If from 5 times a certain number 19 f is subtracted, and the remainder is 18y^^, what is the number.'* 61. I sold my watch for $72, which was f more than I gave for it; what did it cost me ? 62. Bought a horse and saddle for $75, giving | as much fot the saddle as for the horse ; what was the cost of each? 63. A boy, being asked the age of his dog, replied, " If ^ of his age be added to his age, the sum will be 13^ years." What was his age ? 64. Being asked the age of his father, he said, "If 12 years were added to ^ of his age, the sum would equal ^ of his age." What was his age ? Ans. 48 years. 65. Being asked his own age, he answered, *' If 2 years were added to | of my age, the sum would equal ^ of my age.'* What was his age ? Ans. 1 6 years. 66. A can build a wall in 3 days, and B can do the same work in 4 days. What part of the work can each do in one day ? What part can both do in one day? In how many days can both do it working together ? Ans. If days. 67. C can do a piece of work in 5 days, and D in 8 days. What time will be required for both to do it ? Ans. 3jJj days. 96 COMMON FRACTIONS. 68f If E can do the same work in 7 days, how long would be required for C, D, and E, to do it wprking together ? Ans. 2yL8_ days. 69t If A, B, and C can do a piece of work in 6 days, and A and B can do the same work in 8 days, in what time can C do it alone ? Ans. 24 days. 70.* Shipped to Havre 2000 bbls. of flour, which I sold at $7| per bhl. ; received in return oOOf hhds. of wine, worth $21i per ■ hhd. ; what sum is still due me ? 7 It A merchant owned f of a cargo of teas, the whole cargo worth $65000 ; he sells f of his share for $8583.331 y does he gain or lose, and how much ? 72* From a tank containing 184 gallons of water, 20|^ gallons were drawn out ; if | of what then remained was equal to | cf what afterwards rained in, how much rained in? How much did the tank then contain? Ans, 236|^- gallons. 73.* I pay $700 for a piece of land ; cut 52^ cords of wood from it, which I sell at $5.40 a cord ; I pay $1^ a cord for cut- ting and hauling the wood, and $10 for surveying the land ; I divide 3 acres of it into house lots of | acre each ; 4 of these I sell at $175 each, and the rest at $162.50 per lot. Eeserving 2 acres for myself, valued at $300, I sell the remainder of the land for $600, what do I gain ? Ans. $2388.18f. 74t Messrs. B, D, W, and S, built a drain together, each agreeing to pay his proportion of whatever he occupied. B oc- cupied 20 feet alone, B and D 22 feet, B, D, and W, 140 feet. B, D, W, and S, 18 feet. The drain was built at a cost of 33^ cents per foot ; what was each person's share of the cost ? Note. — B's share == 20 X 33 J -f- 22x_3_M 4- lAJi^AM + LSx^^M, Ans. B, $27.38| ; D, $20.72| ; W, $17.05|- ; S, $1.50. 75t A, B, C, and D, hired a team together in Boston for a journey north, each agreeing to share the expense for the dis- tance he rode. At Reading, 14 miles from Boston, A got out ; at Andover, 8 miles further, B got out ; at Lawrence, 4 miles further, C left, and D went on alone 8 miles to Haverhill. Re- GENERAL REVIEW. 9V turning, he took up C, B, and A, where he left them, and all rode into Boston. They paid $8.50 for the use of the team ; what was each one's share ? Note. — The distance from Boston to Haverhill is 34 miles ; the price for I mile out and back is $ ^-^£- = $.25 ; D's share is JLA x 2_5 _^ _8_x^ ^ 1.x 2^ + 8X25. Am, A, 6.87^ ; B, $1.54^; C, $2.04^ ; D, $4.04^. 14:0« General Review, No. 3. 1* What are the prime factors of 420 ? 2, Divide 15 X 7 X 12 X «, by 21 X 10 X 3 X 4. S, What is the greatest common divisor of 21, 84, and 51 ? 4. What is the least common multiple of 42, 9, 14, and 12 ? 5. Reduce ^f f and -^-f^ to their lowest terms, 6. Reduce 25 4f to an improper fraction, 7. Reduce ^|^ to a mixed number. 8. Reduce y\ of -^^ of f ^ of 6| to a simple fraction. 9. Reduce |, |, and f , to a common denominator. 10. Reduce ^f , ^f , and 8f , to the least common denominator, 11. Addfof^, ij, and9^. 12. Add 15|, 3^, and 25^. 13. From ^f of f^- take y\, 14. Subtract S^^ from 10/^. 15. Multiply y4 by 3|. 16. Divideloftkby lOf 17. Change _I_j -f-, -f, and— to simple fractions. 18.* Whaipartof 8f is2i? 19! What is the greatest common divisor of ^S f, and 2f ? 20* What is the least common multiple of ^*y? fV? ^^^ /f • 21* How many fourths of | of 40 in 3f X I -^- f off*? I^ For changes, see Key, i8 COMPOUND DENOMINATE NUMBERS. COMPOUND DENOMINATE NUMBERS. 150 • Numbers are either Simple or Compound. 151* A Simple Ntunber is a number expressed in unitj of one denomination ; as, 5 hoolcs^ 7 pens. 1«S3. A Compound Number is a number expressed in units of two or more denominations, but of the same nature ; as, 5 pounds 6 ounces of sugar ^ 3 years 2 months 4 days of time, 103. Keduction is the process of changing the denomi- nation of numbers without altering their value. 154* Reduction Descending is the process of changing numbers to numbers of equal value in lower dentMninations ; thus, 1 dollar =100 cents. \55^ Keduction Ascending is the process of changing numbers to numbers of equal value in higher denominations; thus, 100 cents z=z 1 dollar. 1«>6* Compound numbers express Currency, Weight, and Measure. Currency. Every nation has its own currency. That of the United States has already been given (Art. 68), bnt the table will be inserted here for the sake of uniformity. 157, Federal Money. The denominations are eagles, dollars, dimes, cents, and mills. The legal coins in circulation are as follows : Gold. Silver. Double Eagle = $20.00. Dollar — $1.00. Eagle =: 10.00. Half Dollar = .50. Half Eagle =: 5.00. ' Quarter Dollar =z .25. Quarter Eagle =: 2.50. Dime = .10. Three Dollar piece =: 3.00. Half Dime = .05. One Dollar piece = 1.00. Three Cent piece z=: ,03. Copper and nickel Cent, and Two Cent piece*. COMPOUND DENOMINATE NUMBERS. 99 Note. — The gold coin is hardened by an alloy of ^ copper and silver (the silver not to exceed the copper). The silver coin is hardened by ^ copper. The cent coined since 1856 has 88 parts of copper to 12 of nickel. The two-cent piece, coined 1864, has 95 parts copper to 5 of tin and zinc. Table. 10 mills (m.) z=: 1 cent, marked c. or ct. 10 c. =r 1 dime, ^* d. 10 d. =11 dollar, " $. $10. =1 eagle, « E. Note. — Mill is derived from the Latin miller one thousand, because 1000 mills = 1 dollar, the unit of computation,- cent from Latin centum, one hundred, because 100 cents = 1 dollar; dime from the French dime, a tenth, as a dime is one tenth of a dollar ; dollar from the German thaler, dollar, dollars having been first coined in Germany. Exercises. 1. Write 3 E. $2. 7 d. 5 c. 2 m. as it is usually written. Ans. $32,752. 2. Write 162 E. $8. 3 d. 9 c. 8 m. as it is usually written. ♦ Ans, $1628.398. Write in the same manner, 3. 128 E. 3 d. 8 m. 4. 19 E. $6. 3 c. 2 m. 5. 68 E. $8. 2 m. 6. $7. 2 c. 5 m. 7. $5. 6 d. 8 c. 3 m. 8. 3984 E. 7 d. 4 c. 8 m. 9. Add the answers of the last six examples, and give the amount in mills. Ans, 42,017,798 mill^ 158, English Money. The denominations are pounds, sMlUngs^ pence^ Q,nd farthings. Table. 4 farthings (qr. or far.) =: 1 penny, marked d. 12 d, =1 shilling, " s. 20 s. =1 pound, " £. Note. — The guinea of 21 s., and the crown of 5 s., are also used. Th^ coin which represents the £ value is called a sovereign. 12 860 d. 4 lt)(> REDUCTION ASCENDING. 109, Eeduction Descendino. Xll. Ex. Reduce 3 ^ 11 s. 8 d. 2 far. to farthings. Operation. ^^ 20 s. m 1 £, we shall have 20 times 3 £ 11 s. 8 d. 2 far. ^^ many s. as £. (20 X 3 ) s. =: 60 s. ; 20 60 s. + 11 s. = 71 s. As 12 d. zz: 1 s., we j~ shall have 12 times as many d. as s. ; (12 X 71) d. = 852 d. ; 852 d. + 8 d. = 860 d. As 4 far. =i 1 d., we shall have 4 times as many far. as d. ; (4 X 860) far. = 3440 far; 3440 far. + 2 far. = 3442 farthings. Hence 3442 farthings, Ans. the EuLE FOR Reduction Descending. Multiply the number of the highest denomination by the number which it takes of the next lower denomination to make one of that higher, and to the product add the given number of the next lower denomination. Multiply that sum in like manner, and thus proceed till the number is reduced to the required denomination. ' Examples. 1. Reduce 7 £ 8 s. 3 d. 3 far. to farthings. Ans. 7119 far. ' 2. Reduce 30 £ 2 s. d. 2 far. to farthings. Ans. 28898 far. 3. Reduce 8 £ s. 3 d. to farthings. 4. Reduce 9 s. Id. 2 far. to farthings. 5. Reduce 368 £ 17 s. 2 d. to pence. 6. Reduce 25 crowns 3 s. 2 d. to farthiiags. 7. Reduce 43 crowns 4 s. 8 d. to pence. 8. Reduce 209 guineas to pence. 9. What will be the number of farthing candles that may be bought for 2 s. 6 d.? 160. Reduction Ascending. III. Ex. Reduce 3579 farthings to an equivalent value in higher denominations. ^ REDUCTION. 101 OPEftATioN. As 4 qr. = 1 d., we shaU have 1 1 juuia ^g many pence as farthings, or 894 d. 12 ) 894 d. + 3 qr. and 3 qr. remaining ; as 12 d. =r 1 s., 20 ) 74 s. ■+- 6 d. y^Q gjiall have ^ as many shillings 3 £ 14 s. 6 d. 3 qr., Ans. as pence, or 74 s. and 6 d. remain- ing. As 20 s. 1= 1 £, we shall have ^ as many £ as s., or 3 £ and 14 s. remaining, making the entire result 3£ 14 s. 6 d. 3 qr. Hence the Rule for Eeduction Ascending. Divide the given number by the number which it takes of its denomination to equal one of the next higher, and note the remainder. Divide the quotient thus obtained as before, and thus -proceed till the required denomination is attained. The last quotient, with the several remainders, will be the required result. Proof. As reduction ascending is the converse of reduction descending, either process may be proved by the other. Examples. 1. Reduce 3681 farthings to an equivalent value in higher denominations. Ans. 3 £ 16 s. 8d. 1 qr. In the same manner reduce, 2. 36875 farthings. Ans. 38 £ 8 s. 2 d, 3 far. 3. 4328 pence. Ans, 18 £ s. 8. 4. 39818 shillings. o. 86347 farthings. 6. 298721 farthings. 161. Comparison of English and Federal Currency. 1 £ = $4.84. How many % in 1. 36 £? ^?is. $174.24 2. 49 £? 3. 64^ £? How many £ in 4. $39.43 ? Ans. 8/3^ £. 5. $43.76? 6. $78.39 ? lot COMPOUND DENOMmAVE NUMBERS. WEIGHT. 163. Troy Weight. Gold, silver, and precious stones are weighed by this system. The denominations are pounds, ounces, pennyweights, and grains. Table. 24 grains (gr.) z=z 1 pennyweight, marked pwt. 20 pwt. z=z 1 ounce, " oz. 12 oz. n: 1 pound, " lb. III. Ex. Reduce 2 lb. 9 oz. 18 pwt. 3 gr. to grains. Operation. 2 lb. 9 oz. 18 pwt. 3 gr. 12 33 oz. 20 678 pwt 24 2715 1356 III. Ex. Reduce 16275 gr. to numbers of higher denomina- tions. Operation. 24 ) 16275 gr. 2|0) 67 1 8 pwt. -I- 3 gr. 12 ) 33 oz. + 18 pwt. 2 1b.+9oz. Ans. 2 lb. 9 oz. 18 pwt. 3 gr. 16275 gr., Ans, Examples. 1. Reduce 18 lb. 11 oz. 5 pwt. 17 gr. to grains. Ans. 109097 gr.- 2. Reduce 48 lb. 2 oz. pwt. 3 gr. to grains. 3. Reduce 1 oz. 23 gr. to grains. 4. Reduce 3681 lb. 9 oz. 1 pwt. to pennyweights. Reduce to equivalent values in higher denominations, 5. 928641 pwts. Ans. 3869 lb. 4 oz. 1 pwt. 6. 3786541 grs. 7. 9042028 grs. 8. What is the value of 2 lb. 8 oz. of gold, at $16 00 an ounce ? REDUCTION. 103 9. What is the value of 1 lb. 3 oz. 7 pwt. of gold, at 4 cents per grain ? 10. What will 20 silver dollars weigh, each dollar weighing 4121 grains.'^ 163. Apothecaries* Weight. Apothecaries use this weight for mixing medicines ; but thejT buy, and generally sell, by Avoirdupois weight. The denominations are pouyids, omices, drachms, scruples, and grains. Table. 20 grains (gr.) = 1 scruple, marked se. or 3. 3 9 z=l drachm, " dr. or 5- 8 5 ^=1 ounce, " oz. or § - 12 .§ z= 1 pound, " lb. or fc. III. Ex. Reduce 2 ife. 3 S, 2 5, 1 9, 5 gr. to grains. Operation. 21b 12 27 S 8 218 5 3 655 3 20 III. Ex. Reduce 68321 grains to numbers of higher denominations. Operation. 210)6832J1 gr. 3)3416 9 4-lgr. 8)1138 3 + 2 9. 12)142 14-2 3. 11 lb. 4- 10 §. Jbw. lllb.4-10B,23,2 9, Igr. 13105 gr., Ans. Examples. 1. Reduce 5 lb. 7 i , 75, 2 9, 12 gr. to grains. Ans. 32632 gr. 2. Reduce 3 lb. S , 7 5, 1 9, 9 gr. to grains. 3. Reduce 258481 grains to pounds, ounces, &c. Ans, 44 lb. 10 S, 4 3,0 9, 1 gr. 4. Reduce 36845 9 to pounds, ounces, &c. 104 COMPOUND DENOMINATE NUMBERS. 5. Eeduce 987326 gr. to pounds^ ounces, &c. 6. Reduce 28 ft. 3 S, 1 5, 2 9, 5 gr. to grains. Avoirdupois Weight. ling almost all articles, except 164. This weight is used for wei| gold, silver, and precious stones. The denominations are tons, hundred weighty quarters y pounds^ ounces, and drams. Table. 16 drams (dr.)=:l ounce, marked oz. 16 02. :=1 pound, " lb. 25 Ibw =1 quarter, " q^. 4 qr. z=i\ hundred weight, " ewt. 20 ewt. =1 ton, « T. Note. — The long ton of 2240 lbs., which gives 2S lbs. to the qr., is sometimes used for weighing gross articles, as iron and coal, and is the ton recognized by the United States Government. III. Ex. Reduce 2 ewt. 3 qr* 8 lbs. to pounds. Operation. 2 ewt. 3 qrs.Slbs. 4 11 qr. 25 63 22 283 1b.,^ra*. III. Ex.— Reduce 186421 dr. to numbers of higher de- nominations. Operation. 16)186421 dr. 16)11651 oz.4-5 dr. 25)72?'lb. + 3oz. 4) 29 qr. 4-3 lb. 7 ewt. + 1 qr. Ans. 7 ewt. 1 qr. 3 \h. 3 oz. 5 dr. Examples. 1. Reduce 5 ewt. 3 qr. 24 lbs. to pounds. 2. Reduce 23 T. 4 lbs. to ounces. Ans. 599 lb& Reduce to equivalent values in higher denominations. 8. 9328 lbs. Ans. 4 T. 13 ewt. 1 qr. 3 lb. i. 36842 oz. 5. 193256 lbs. 6. 8236548 dr. 7. 9654321 dr. EEDUCTION. 105 8. Reduce 4 T. 3 cwt. 2 qr. lb. 8 oz. to ounces. 9. How many ounces in 1 cwt. ? in 1 T. ? 10. How many pounds in one ton ? 11. How many more pounds in a long ton than in a short ton ? 12. At the rate of 3 lb. a day, how many hundred weight of flour will a family consume in a year, or 365 days? 165. Comparison of Weights. Troy. Apotii. Av. lib. = lib. r^ Wlb. 1 oz. == 1 § :z= Iffoz. Igr. = Igr. =z T^oT^b. 000 gr. = 7000 gr. Examples. ^^^ 1 lb. 1. Reduce 364 lbs. Troy to Avoirdupois weight. Ans. 299tV^ lbs. Av. 2. Reduce 36 lbs. Troy to Avoirdupois weight. 3. Reduce 5 lbs. Avoirdupois to grains in Apothecaries' weight ; to units of higher denominations. Ans. 6 lb. § , 7 3, 1 B. 4. Reduce 375 lbs. Avoirdupois to Apothecaries* weight. 5. Reduce 73 lbs. Avoirdupois to Troy weight. Measures of Extension. 166. Long Measure. The denominations are miles, furlongs, rods, yards, feet, inches and lines. Table. 12 lines (1.) z= 1 inch, marked in. 12 in. ^Ifoot, « ft. 3 ft. =1 yard, « yd. 5 J yd. or 16^ ft. := 1 rod, " r. or rd. 40 r. zzz. 1 furlong, " f. or fur. 8 f. =1 mile, " m. 69^ miles nearly = 1 degree (°) of longitude at the Equator, 360 of which degrees z=. the distance round the earth. 3 miles zzz 1 land league. 1 mile = 320 rods = 5280 feet. 106 COMPOUND DENOMINATE NUMBERS. III. Ex. Reduce 2 m. 5. f. I III. Ex. Reduce 1543514 13 r. 4 yd. 2 ft. to inches. I inches to miles, furlongs, etc. Operation. m. f. r. y. ft. 2. 5.* 13. 4. 2. 8 21 f. 40 853 r. 426^ 4269 4695i 3 yd. 14088^ ft. 169062 in,, Ans. Operation. 12 ) 1543514 in. 3 ) 128626 ft. + 2 in. 5^ == J^ ) 42875 y. + 1 ft. 2 11 ) 85750 halves of yd. 410 ) 77915 r. + I yd. -| yds. = 2 yd. 1 ft. 6 in. 8 ) 194 f. + 35 r. 24 m. + 2 f. 24 m. 2. f. 35 r. 2 yd. 1 ft. 2 in. 1 ft. 6 in. Ans. 24 m. 2 f. 35 r. 2 yd. 2 ft. 8 in. Examples. 1. Reduce 3 m. 7 f. 14 r. to yards. A7is. 6897 yds. 2. Reduce 3 f. 11 r. 2 yd. 1 ft. 7 in. to inches. Ans. 26029 in. 3. Reduce 1590 inches to rods, etc. Ans. 8 rd. yd. ft. 6 in. 4. Reduce 5 m. 6 f. 3 r. 3 yd. 2 ft. 5 in. to lines. 5. Reduce 16906 inches to numbers of higher denominations.^ 6. Reduce 1291968 lines to miles, furlongs, etc. 7. How many miles round the earth ? 8. How many miles through the earth from pole to pole, the distance being 41704788 feet ? 9. Find the cost per mile for grading a road, at 90 cents per rod. Ans. $288. 10. What will it cost to fence both sides of a road, 26 r. 2 yd. long, at $.65 per yd. ? - Ans. $188.50. 11. How many furrows, each, 10 in. wide, will be made in ploughing a lot of land lengthwise, which is 6 r. 1 ft. wide ? REDUCTION. 107 167« Surveyors' Measure. The denominations are miles, chains, rods, links, and inches. Table. 7y^5% inches z=z 1 link, marked 1. 25 1. =1 rod, « r. 4 r. zz: 1 chain, " ch. 80 ch. = 1 mile, « m. 1 chain = 4 rods =: 66 feet = 100 links =i 792 inches. Note. — Rods are seldom used by surveyors, the distances being gen»- lally taken in chains and links. III. Ex. Reduce 4 m. 75 1 III. Ex. Reduce 763218 ch. 32 1. to links. 1 links to miles, chains, etc Operation. Operatiok, 4 m. 75 eh. 32 L 100 ) 763218 ; > 80 810)76312 ch-hl. 395 ch. 100 95:m. + 32ch. 39532 1., Ans, Am, 95 m. 32 ch. 18 1. Examples. 1. Reduce 3 m. 35 ch. 8 1. to links. Ans, 27508 1. 2. Reduce 5 m. 78 ch. 2 1. 5 in. to inches. 3. 13845 links to miles, chains, etc. Ans. 1 m. 58 ch. 45 1. 4. 259248 inches to miles, chains, etc. 5. In 1 m. 46 ch. 2 r. how many rods ? 6. In 9584 feet, how many chains ? 168* Mariners' Measure. The denominations for short distances are cable-lengths, fathoms, and feet. Table. 6 feet (ft.) z=z 1 fathom, marked fath. 120 fath. r=: 1 cable-length, marked c. 1. . 7|^ c. 1. zzz\ common mile, " m. Longer distances are estimated in nautical or geographical miles, each mile being ^^ of 2^ degree me^ured on a great circle 108 COMPOUND DENOMINATE NUMBERS. of the earth, and averaging 6086.34 ft.* or 1.15 -j- common milec 3 nautical miles = 1 sea league. Examples. 1. How many feet in 7 c. 1. 32 fath. ? Am. 5232 ft 2. How many feet in 5 c. 1. 4 ft. ? 3. How many cable-lengths in 672 fath. ? Ans. 5 c. 1. 72 fath. J. Keduce 3684 feet to units of higher denominations. 169. Cloth Measure. Cloth is measured by its length, without regard to its width, ^iie yard is considered the unit of measure, and is divided into halves, quarters, eighths, and sixteenths. 170. Square Measure. This measure is used for determining the area or content* of surfaces. The denominations are square miles, acres, roods, square rods, square yards, square feet, and square inches. Table. 144 square in. (sq. in.) ■=. 1 square foot, marked sq. ft 9 sq. ft. z=z\ square yard, " sq. yd. 30J sq. yd. or 272|^ sq. rt. =r 1 square rod, " sq. r. 40 sq. r. =1 rood, " R. 4 R. z=il acre, " A. 640 A. rz: 1 square mile, " sq. m. III. Ex. Reduce 2 sq. m. 87 A. 3 R. 19 r. to rods. . Operation. 2 sq. m. 87 A. 3 R. 19 r. 640 1367 A. 4 5471 R. 40 218859 r., Ans, Topographical Bureau, Washington, 2864. REDUCTION. 109 III. Ex. Reduce 386060 sq. in. to square rods, yards, &c. Operation. 144 ) 386060 sq. in. 9 ) 2680 sq. ft. + 140 sq. in. 30J ) 297 sq. yd. + 7 sq. ft. 4 4 121) 1188 9 sq. rd. + 24| sq. yd. I sq. yd. — 6 sq. ft. 108 sq. in. (Art. 198.) 9 sq. rd. 24 sq. yd. 6 sq. ft 108 sq. in. 7 sq. ft. 140 sq. in. Ans. 9 sq. rd. 25 sq. yd. 5 sq. ft. 104 sq. in. r. Reduce 3 sq. m. 35 A. to square yards. Ans. 9462200 yd. 2. Reduce 19 A. 2 R. 5 sq. rd. to square inches. Reduce to numbers of higher denominations, 3. 9687 sq. rd. Ans. 60 A. 2 R. 7 r. 4. 5652 sq. yd. 5. 32865 sq. ft. 6. 84791 sq. in. 7. 932485 sq. in. 8. What are 2 A. 3 R. 15 r. 3 y. 8 ft. of land worth at 5 cents a foot ? 171, A rectangle is a figure whose opposite sides are equal, and whose angles are right angles. (Art. 191.) 17^. A square is a rectangle whose sides are all equal. 173* The area of a rectangle is found by rnultiplying its length hy its breadth. Illustration I. Suppose the length of the figure A, B, C, D, to be 4 inches, and its breadth 3 inches. By dividing the line B C into 4 equal parts, and C D into 3 equal parts, and drawing lines from the points of division as in the figure, it will readily be seen that the entire figure- is divided into 4 X 3, or 12 A D I \ c 1 inch. equal parts, each part containing 1 square inch. 110 COMPOLND DENOMIl^ATE NUMBERS. Illustration II. A figure 1 in. long and 1 in. wide contains 1 eq. in. A figure 4 inches long and 1 inch wide must contain 4 times as many sq. inches, or 4 sq. inches. A figure 4 inches long and 3 inches wide must contain 3 times as many sq. inches as if it were only 1 inch wide, or 4 X 3 sq. inches. The area of a rectangle being found by multiplying its length by its breadth, it follows that When the area and one dimension of any rectangle are given, the other dimension may be found by dividing the area by the given dimension. Note. — ^In performing this operation, express the dividend in the superficial denomination corresponding to the linear denomination of the divisor ; that is, if the divisor is expressed in feet, the dividend must be expressed in square feet ; if in yards, the dividend must be expressed in square yards, &c. 174, Examples. 1. If one side of a rectangular field is 16 r. 7 ft., and the other 12 r. 5 ft., how many square feet does it contain? 2. If one side of a square field is 4 r. 8 ft., how many square feet does it contain ? 3. If a rectangular field measures 24 r. 2 fit. in length, and 17 r. 4 yd. in breadth, how many square yards does it contain ? 4. If a floor contains 36 square yards, and its length is 18 ft., what is its width ? Ans. 18 ft, 5. If a ceiling contains 30 6|^ sq. ft., and its width is 17^ ft., what is its length ? 6. A garden containing f of an acre measures on one side 192f feet ; required the length of the other side. 7* How many square feet and inches does the top of a table contain, which measures 3 ft. 2^ in. by 4 ft. 8 in. ? 8. How many square yards of carpeting will be required to cover a floor 17 feet in length by 13 feet in width ? 9. What is the cost of oil-cloth to cover a floor 12 feet by 16,^ feet, at 75 cents per square yard ? DEDUCTION. Ill 175J. Solid or Cubic Measure. This measure is used in finding the contents of solid bodies or space, i. e., of anything that has length, breadth, and thickness; height or depth. The dimensions are cubic yards, cubic feet, and cubic inches. Table. 1728 cubic inches (cu. in.) = 1 cubic foot, marked cu. ft. 27 cubic feet, :=: 1 cubic yard, " cu. yd. Ncra. — The denomination ton is sometimes used, but its value is variable, a greater number of feet being assigned to the ton for light bulky articles than for the heavier. In measuring firewood and some other merchandise, the de- nomination cord is used. A pile of wood 4 feet wide, 4 feet high, and 8 feet long, contains 1 cord. A pile 4 feet wide, 4 feet high, and 1 foot long, contains 1 cord foot. Hence, 16 cu. ft. 8 cd. ft. or 128 cu. ft. rz: 1 cord foot, marked cd. ft. = 1 cord, " cd. 1 cord. led. ft. J cord, Examples. 1. In 3 cu. yd. 18 cu. ft. 136 cu. in. how many inches ? Arts. 171,208 cu. in. 2. Reduce 5 cu. yds. 8 cu. ft. 736 cu. in. to inches. 3. Reduce 368742 cu. in. to cubic yards, feet, &c. Ans. 7 yd. 24 ft. 678 in. 4. Reduce 3427948 cu. in. to cubic yards, feet, &c. 5. How many cord feet in 36 c. 5 cd. ft. ? 6. How many cords in 54328 cu. ft. ? 112 COMPOUND DENOMINATE NUMBERS. 176* A solid bounded by six equal squares is called a Cube. The squares are called the faces of the cube, and, together, make its surface. The bound- ing lines are called edges. If its edges are 1 inch long, it contains 1 cu. in. ; if 1 foot long, it contains 1 cu. ft., &c. 177, A solid that is bounded by rectangles is called a Kectangular Parallelepiped ; rectangular, because its faces are rectangles, and parallelopiped, because its opposite faces are parallel. 178, The Solidity of a Parallelopiped equals the product of its three dimensions. Illustration. Let the figure A B represent a parallelopiped 4 . feet long, 2 feet wide, and 3 feet I high. If it is 4 feet long and 2 feet wide; its lower face or base must contain 4 X 2 zz: 8 square feet. If upon these square feet the solid ex- tends 1 foot high, it will contain 8 cubic feet resting upon the base. But the solid is 3 feet high, and must, therefore, contain three times as many cubic feet as if it were only 1 foot high, or 3 X 4 X 2 cubic feet = 24 cubic feet. Examples. 1. If a solid is 3 ft. long, 5 ft. wide, and 2 ft. high, how many cubic feet does it contain ? Ans. 30 cu. ft. 2. How many cubic inches in a block 3 in. wide, 4 in. high, and 1 ft. 2 in. long? Ans. 168 cu. in. 3. How many cords in a woodpile 40 ft. long, 4 ft. wide, and 4 ft. high? 179, If the solidity of a parallelopiped equals the product of its three dimensions, it follows that When the solid contents and two dimensions are given, the third can he found hy dividing the contents hy the product of the two given dimensions. When the solid contents of a block and the area of its base are given, how do you find its height ? When its contents and height are given, how can you find the area of its base ? 4 feet. REDUCTION. 113 4. How high must a box be made, to contain 24 cu. ft., the length of the box being 4 ft. and its width 3 ft. ? Ans. 2 ft. 5. How high must it be, if its length is 8 ft. and its breadth 3 ft. ? 6. If its height is 2f feet, what must be the area of its base ? 7. How long must a pile of wood be, which is 4 ft. wide, 3 ft. 6 in. high, to contain a cord ? 8. There are 144 square inches on one side of a block con- taining a cubic foot ; what is the length of the edge of the block ? 9. There being 112^ cubic feet in a stick of timber which is 1 J feet square at the end, what is the length ? Ans. 50 ft. Measures of Capacity. 180* Liquid Measure. The denominations are gallons, quarts, pints, and gills. Table. 4 gills (gi.) =: 1 pint, marked pt. 2 pts. z=z 1 quart, " qt. 4 qts. , = 1 gallon, " gall. Note. — The denominations tierce, barrel, hogshead, pipe, butt, and ton, are sometimes used, but their size is variable. Barrels generally contain 31-^ or 32 gall, ; hogsheads, 63 gall. Casks are generally gauged and marked accordingly. They are called hogsheads, pipes, butts, or tuns, without distinction. Examples. 1. Reduce 3 gall. 3 qt. 1 pt. 2 gi. to gills. Ans. 126 gi. 2. Reduce 5 gall. 1 qt. pt. 3 gi. to gills. 3. Reduce 23684 gills to gallons. Ans. 740 gall. 1 pt. 4. Reduce 984324 gills to hogsheads. 5. What will 27 gall. 3 qt. of milk cost at 4 cents per qt. ? 181. Dry Measure. The denominations are bushels, pecks, quarts, pints, and giUs, Table. 4 gills (gi.) zzz. 1 pint, marked pt. 2 pts. zzz 1 quart, '^ qt. 8 qts. zzz 1 peck, " pk. 4 pks. = 1 bushel, " bu. 8 114 COMPOUND DENOMINATE NUMBERS. Examples. 1. Redu«e 5 bu. 3 pk. 3 qt. 1 pt. to pints. Ans. 375 pt. 2. Reduce 2641 pt. to bu., etc. Ans. 41 bu. 1 pk. qt. 1 pt. 3. Reduce 10 bu. 1 pk. 2 qt. pt. 3 gi. to gills. 4. Reduce 8765432 gi. to bu., etc. 5. "What will 4 bu. 1 pk. 2 qt. of cherries cost at 8 cts. per quart? 6. Sold 3 bu. 3 pk. 5 qt. of peaches for $7.50 ; what did receive per quart ? 183. Comparison of Liquid and Dry Measures. Liq, Meas. Dry Meas. Cu. in. 1 qt. =z 57|. 1 gall. =: 231. 1 qt. r= 67f 1 bu. =: 2150f. Examples. 1. I have a dish that contains 2 cu. ft. ; how many quarts of blackberries will it hold? Ans. 51^. 2. How many quarts of water? Ans. 59 ff qt. -3. How many gallons of water will a cistern hold that is 3 ft. long, 3 ft. wide, and 2^ ft. high? 4. How many bushels of apples can be put into a bin 8 ft. long, 3 ft. 2 in. wide, and 2 ft. high? Circular or Angular Measure. 183. This measure is used principally in astronomy, geogra- 'phy, navigation, and surveying. 184, A Circle is a plane surface bounded by a line, every part of which is equally distant from \ a point within called the centre. 185. The bounding line is called the Circum- ference of the circle. Any part of the circumfer- ence is called an Arc. Fig- 2. 186. A straight line passing from the centre of the circle to the circumference, is called a Radius (plural, radii). 187. A straight line passing from one point in the circumference, through the centre, to an opposite yoint, is called a Diameter. REDUCTION. 115 188. The circumference of any circle is supposed to be divided into 360 equal parts, called Degrees, each degree into 60 Minutes, and each minute into 60 Seconds. Table. 60 seconds {") z=z 1 minute, marked ', 60' = 1 degree, " «. 360 ° z= 1 circumference, " circ. 189. A Semi-circumference is half a circumference, a Quadrant one fourth, and a Sextant one sixth. A Sign, used only in astronomy, equals 30°. Fig. 3. 190 • An Angle is the opening between '^ two lines which meet each other. The point of meeting is called the Vertex of the angle. The angle in the annexed figure may be read^ " the angle a h c," or simply " the angle h." An angle is measured by that part of the circum- ference of a circle included between its sides, the centre of the circle being at the vertex of the angle ; thus, rig. 4. In fig. 4, the angle defis measured by the arc mn ; that is, if the arc mn contains 70°, the angle defia aw angle of 70°. 191. An angle which includes or |- of a circumference, is a Right Angle, the sides of which are said to be perpendicular to each other ; in fig. 4, the angle g eh is a. right angle. An angle greater than a right angle is an Obtuse angle. An angle less than a right angle is an Acute angle; hed is a,n acute angle and ^ e c? is an obtuse angle. Note. — As arcs are measurements of angles, the table for angular measure is the same as the table for circular meastire. 199. Examples. 1. Reduce 148° 54' 18 ' to seconds. Ans, 536058 '. 2. Reduce 354° 0' 16" to seconds. 116 COMPOtTND DENOMINATE NUMBERS. 3. Eeduce 53684" to numbers of higher denominations. Ans. 14° 54' 44". 4. Reduce 359° 59' 59'' to seconds. 5. Reduce 1 quadrant to seconds. 6. How many seconds in 1 sextant ? 7. How many minutes in a sign ? 8. Reduce 35467" to numbers of higher denominations. Time Measure.- 193* The length of an Astronomical or Sidereal Day is the time the earth takes to turn once upon its axis; the length of a Solar Day is the time the earth takes to turn so as to bring the sun to the same meridian again. The solar day is divided into 24 hours, each hour into 60 minutes, and each min- ute into 60 seconds. The denominations of time are centuries, years, months, weeks, days, hours, minutes and seconds. Table. 52 w. 194* The time which the earth takes to revolve around the sun is 365 d. 5 h. 48 m. 50 s. nearly. The common year (365 days) thus loses nearly one day in 4 years. Hence the leap year of 366 days was established, which occurs once in 4 years. But this adds too much by about 11^ m. a year, which in 100 years amounts to nearly 18§ h. To balance this error, every 100th year is not regarded as a leap year. But this drops too much by a little more than 5] h., which in 4 centuries amounts to nearly 1 d. Hence every four-hundredth year is a leap year. This leaves an error which is less than 1 d. in 3600 years. Hence the 60 seconds (8.) z=z 1 minute, marked m. 60 m. := 1 hour. « h. 24 h. := 1 day, d. 7d. = 1 week, « -m d. or 365 d. = 1 common year, " c. y. 366 d. r= 1 leap year, 1. y. 365^ d. :=: 1 Julian year, " J. y. 100 y. =i 1 century, C. REDUCTION. 117 Rule for ascertaining when any year is a leap year. When the number denoting the year is divisible by 4, and not by 100, it is a leap year ; and any year that is divisible by 400 is a lea/p year, ] 0«5, A year is divided into four seasons, of three calendar months each, and commences with January, the second winter month. The succession of th^ seasons , quarters and months, and the number of days in each month, are shown by the following diagram : Quarter Quarter. «• Thirty days hath September, April, June, and November ; All the rest have thirty- one, Except February alone, To which we twenty-eight assign, Till leap year gives it twenty-nine." Note. In the following examples, common and leap years are u»der« Btood unless the Julian is specified. * Leap year. 29 d. 118 COMPOUND DENOMINATE NUMBERS. III. Ex. Reduce 2 y. 7 w. 4 d. 4 h. 33 m. to minutes. Operation. 7 w. 4 d. = 53 d. 2 y. 53 d. 4 h. 33 m. 365 783 d. 24 3132 1566 18796 h. 60 1127793 m. III. Ex. Reduce 5387294 minutes to numbers of higher denominations. Opkration. 610)53872914 m. 24 ) 89788 h. + 14 m. 365)3741 d. + 4 h. 10 y. + 91 d. 2 89 d. As 2, at least, of the 10 years must be leap years, 2 days should be taken from the 91 days remain- ing, which leaves 89 days. Ans. 10 y. 89d. 4h. 14 m. 196. Examples. 1. Reduce 8 y. 3 w. 19 d. 7 h. to hours, allowing for 2 leap years. Ans. 71095 h. Note. 2 d.-f- 3 w. 19 d.= 42 d., .*. the example may be stated. Reduce 8 c. y. 42 d. 7 h. to hours. 2. Reduce 13 y. 8 w. 2 d. 3 h. 18 m. to minutes, allowing for 3 leap years. 3. Reduce 180739 hours to numbers of higher denominations. Ans. 20 y. 225 d. 19 h. Reduce 5683762 minutes to numbers of higher denomina- 4. lions 5. 6. How many minutes in the 1st century ? Ans. 52594560 m. How many hours in 10 y. 36 d., beginning with Jan. 1st, 1852? Ans. S853Qh. 7. How many seconds in the 3 summer months ? 8. How many days from April 12th, 1831, to May 3d, 1832 ? Note. — From April 12, 1831, to April 12, 1832 = 366 days; to May 3, 21 days more. Ans. 387 days. 9. How many days from Jan. 1st, 1832, to Jan. 1st, 1863 ? 10. How many days from March 1st, 1850, to Jan. 1st, 1864? 11. How many seconds in 10 years, 3^6 minutes, allowing 365| days to the year ? A book formed of sheets folded KEDUCTION. 197. Miscellaneous Table. Numbers. \ 12 units or single things := 1 dozen. 12 dozen z=z 1 gross. 12 gross ==: 1 great gross. 20 units or single things ::== 1 score. Paper. 24 sheets of paper nz 1 quire, 20 quires z= 1 ream. in 2 leaves, is a folio, in 4 leaves, is a quarto, in 8 leaves, is an octavo, in 12 leaves, is a duodecimo or 12mo, in 16 leaves, is a 16mo. in 18 leaves, is an 18mo, in 24 leaves, is a 24mo. in 32 leaves, is a 32mo. ^ in 64 leaves, is a 64mo. Height of Animals. 3 in. = palm. 4 in. :n hand. 9 in. z= span. Capacity. 1 barrel of flour = 196 lbs. 1 barrel of pork z=z 200 lbs. Examples. 1. How many rows of buttons, 6 in a row, are there in a great gross of buttons ? 2. In 3 score and 6 years how many days ? 3. How many sheets of paper in 3 reams, 7 quires, 21 sheets ? 4. How high must a doorway be for a horse to pass freely un- der that is 151 hands high? 5. How many loaves of bread can be made from a barrel of flour, allowing 12| oz. to the loaf? 6. If pork is worth $18.75 a bbl., what is it worth per lb.? 120 COMPOUND DENOMINATE NUMBERS. Suggestion. The pupil may now write from memory and present for inspection, or repeat forward and backward, the table of Federal Money; of English Money; of Troy Weight; of Apothecaries' Weight ; of Avoirdupois Weight ; of Dry Measure ; of Liquid Measure ; of Long Measure ; of Mariners' Measure ; of Surveyors' Measure ; of Square Measure ; of Cubic Measure ; of Circular Measure ; of Time. IS^ For Dictation Exercises in Reduction, see Key. FRACTIONAL APPLICATIONS. 198. Reduction of a Fraction of one Denomination TO Whole Numbers of Lower Denominations. III. Ex., I. Reduce |£ to shillings, &c. Operation. f £ = I of 20 s. r=; ^o s. = 13^ s. 1 s. r= ^ of 12 d. = 4 d. Ans. 13 s. 4 d. III. Ex., II. Reduce f cwt. to quarters, pounds, &c. Opekation. 6 cwt. = f of 4 qr. r=: -\o qr. r= 2f qr. f qr. z= 6 of 25 lb. z=. Ifii lb. = 21f lb. •f lb. = f of 16 oz. — Y- oz- = 6f oz. f oz. = f of 16 dr. = -\6. (Jr. = 13f dr. Ans. 2 qr. 21 lb. 6 oz. 13| dr. Or, expressing the work in an abbreviated form, I cwt. z= Y qr. = 2f qr. 4qr. =:JL5o lb. — 21f lb. f lb. = -V oz. = e^ oz. f oz. z=z -^6. dr. = 13f dr. Hence the Rule. To reduce a fraction of one denomination to whole numbers of lower denominations : Multiply the fraction hy the number which it takes of the next lower denomination to make one of that ; reduce the fraction thus obtained to a whole or mixed number, if possible. If a fraction remain, proceed with it «5 before, and thus continue as far as desired. FRACTIONAL APPLICATIONS. 121 Examples. Reduce to whole numbers of lower denominations, 1. I of 1 £. Ans, 16 s, 8 d, 2. J of 1 lb, Troy, Ans, 10 oz. 10 pwt, 3, t of 1 lb, 4. 1^ of 1°. 5, 1^ of 1 cwt. C. I of 1 c. y, 7. -^2 ^^ 1 gallon, 6, I of 1 bu. 9. ^y of 1 mile. 10. 1^ of 1 furlong. 11. y^j of 1 chain. 12. f of 1 league. 13. ^2 ^^ 1 s^g0 oz. = ^ oz. 7i0Z, = -2j2 0Z, = 3x^;^ lb. = -i|lb., ^ws. Hence the Rule. To reduce whole numbers of lower denominations to the fraction of a higher denomination : Reduce the number of the lowest denomination to a fraction of the next higher. Annex it to the number of that higher denomination^ and change the mixed . number thus obtained to an improper fraction^ Reduce as before, and thus continue as far as desired. 122 COMPOUND DENOMINATE NUMBERS. Examples. Reduce 1. 6 s. 3 d. to the fraction of a £. Ans. ^% £. 2. 3 p. 6 qt. 1^ pt. to the fraction of a bu. Ans. ff bu. 3. 1 qt. pt. 1 gi. to the fraction of a gall. 4. 1 §, 2 3, 2 9 , to the fraction of a lb. 5. 5 cwt. 1 qr. 16 lb. 10§ oz. to the fraction of a T. 6. 6 fur. 2 r. 2 y. 1 ft. to the fraction of a m. 7. 4 y. ft. 4J- in. to the fraction of a r. 8. 2 r. 1 1. to the fraction of a ch. What part of 9. 1 A. is 2 R. 1 r. 24 sq. y. 6 sq. ft. 108 sq. in. ? 10. 1 cu. yd. is 13 cu. ft. 864 cu. in.? 11. 1 cd, is 5 cd. ft. 4 cu. ft. 576 cu. in.? 12. 1 c. y. is 162 d. 5h. 20 m.? 13. 1 1. y. is 146 d. 9 h. 36 m.? 14. 1 J. y. is 350 d. 15 h. 21 m. 3Q s.? 15. If 1 £ is worth $4.84, what is the value of 4 s. 6 d. ? Solution. 4 s. 6 d. =: ^^^ £. 1 £ =2 $4.84, . • . ^ £ — ^^ of $4.84 1= $1,089. What cost 16. 3 pk. 2 qt. of meal at $X^0 a bu. ? 17. 2 qt. 1 pt. of kerosene oil at $.52 a gall. ? 18. 62 lb. 8 oz. soap at $7.50 per cwt. ? 19. 2 g, 1 3, 4 gr., quinine at $4.00 per g? 20. How long will it take a man to travel 9 miles at the rate of 3 m. 6 f. 26 r. 3 yd. 2 ft. an hour? 21. At $60.00 an acre, what cost 2 A. 3 R. 13^ sq. rd.? 22. At $9.00 a ton, what cost 1 T. 5 cwt. 2 qr. 14 lb. of coal ? (Long ton.) 23. At $198 a lb., what cost 10 oz. 10 pwt. 10 gr. of gold? 24. Tlie weight of a cubic foot of water being 62|^ lbs., how many pounds of water will a tank contain which measures 9 ft. 6 in. by 8 ft. 8 in., and is 6 ft. 9 in. deep ? 25. A cubic foot of granite weighs 163 lbs. 5 oz. ; what is the weight of a block 3 ft. 2f in. by 2 ft. 4 in. and 1 ft. 3 in. thick? I^" For Dictation Exercises, see Key, ADDITION. 123 300. Addition. Addition of Compound Numbers is the process of finding a number equal in value to two or more given compound numbers. The process is similar to addition of simple numbers. III. Ex. What is the sum of 3 £ 11 s. 6 d. 3 qr., 4 £ 7 s. 8 d. 2 qr., 7 s. 6 d. 2 qr., and 9 £ 18 s. ? Operatiox. Writing the numbers, pounds under pounds, £. s. d. qr. shillings under shillings, &c., we commence 3 11 6 3 by adding the numbers in the farthings' col- 4 7 8 2 umn, and find the amount = 7 qr. rz: 1 d. + 3 qr. 7 6 2 Writing 3 in the farthings' place, we add the 9 18 Id. with the column of pence, and have for the ■ amount, 21 d. =r 1 s. + 9 d. Writing 9 in the Ans. 18 £ 4 s. 9 d. 3 qr. ^^^^^, ^^^^^^ ^^ ^^^ ^^^ j ^^ ^j^j^ ^^^ shiUings, and have 44 s. =: 2 £ -j- 4 s. Writing 4 in the shillings' place we add the 2 £ with the column of pounds, and have for an answer, 18 £ 4 s. 9 d. 3 qr. Hence the Rule for Addition of Compound Numbers. Write the numbers of like denominations in the same column, and commence in adding with the numbers of the lowest denomination. Divide the amount hy the number it takes of that denomination to make one of the next higher, write the remainder under the column, and add the quotient with the numbers of the next higher denomination. Add the next column in the same manner, and thus continue till all the numbers are added. SOI. Examples, Add the following numbers : CURRENCY. 1. 2. a. $ £. s. d. qr. £. s. d. qr. 49.703 5 8 4 2 206 18 4 3 8.47 7 15 3 3 29 14 9"^ 2 .882 19 2 118 7 10 4.369 16 4 6 13 7 1 Ans. 30 7 2 3 124 -4^*' III. Ex., II. Reduce .16 to a common fraction. 1# If Operatiox. .16z=:-^ziz— z=J^. Ans. 10 10 6' Examples. Reduce the following to common fractions : 11 L .6. Ans. |. 2. .83. Ans. I. a .1881 162 DECIMAL FRACTIONS. 4. .428571. 5. .714285. 6. .2142857. 944, To Reduce Compound Numbers to Decimal Fractions op Higher Denominations. III. Ex., I. Reduce 2 d. 3 qr. to the decimal of a shilling. Operatiox. Since 4 qr. equal 1 d., there will be \ as many d. 4 3.00 qr. as qr., or | d., which equals .75 d. ; this, with the 12 2 75000 d ^ ^* S^^^^^» equals 2.75 d. ; since 12 d. equals 1 -. ■ shilling, there will be ^ as many shillings as d., III. Ex., II. What is the value of 3 rds. 4 yds. 2 ft. in the decimal of a rod ? Operation. gi^ce 3 ft. equal 1 yd., there will be ^ as 2.00000 It. many yds. as feet, or | yds., which equals 4.66666 4- yd. ^g yds., this, with the 4 yds. given, equals 4.6 yds. ; since 54 yds. equals 1 rod, there 9.33333+ half yd. , i !? ^ i 3^4848 + rods, J[n.. ^'^^^ ^^ 5^ °" ^'^ ^' ^^"^ ^'^^^ ^' ^^^" ^"- From the above, we deduce the following Rule. To reduce compound numbers to decimal fractions of higher denominations : Divide the number of the lowest denom-^ ination by what it takes of ifiat denomiiiation to make one of the next higher ; 'place the quotient as a decimal fraction at the right of that higher ; so continue till all the terms are reduced to the denomination required. Examples. 1. Reduce 7 d. 3 qr, to the decimal of a £. A)is. £.03229+. 2. Reduce 3 da. 22 h, 4 m. 48 sec, to the decimal of a week. Ans, .56 wk. 3. Reduce 5 cwt. 3 qr, 10 lb, to the decimal of a ton. 4. Reduce 5 cord ft. 12 cu. feet to the decimal of a cord, 5. Reduce 10 oz. 5 pwt. 12 gr. to the decimal of a pound. 6. Reduce 80 cu, ft. to the decimal of a cord. REDUC110JS 163 7. What is the value of 2 fur. 7 rd. 10 ft. expressed in th« decimal of a mile ? 8. What part of a ream is 15 quires 12 sheets ? 9. What part of an acre is 3 R. 15 rd. 6 yd. 83 ft. ? 10. Reduce 7 S. 8° 5' 38" to the decimal of a great circle. I^^ For Dictation Exercises, see Key. 345. To Reduce Decimal Fractions of Higher De- nominations TO Whole Numbers op Lower Denomina- tions. III. Ex. Reduce .13125 lbs. Troy to oz., &c. Opkration. lb. .13125 Since 12 oz. = 1 lb., there will be 12 times as ]^ many ounces as pounds, =z 1.575 oz. ; since 20 pwt. oz. 1.575 m 1 oz., there will be 20 times as many pwt. as ^ ounces, =z 11.5 pwt. ; since 24 gr. :=. 1 pwt., there pwt. 11.5 will be 24 times as many grains as pwt., rz: 12 gr. ^"^ Ans. 1 oz., 11 pwt., 12 gr. Hence the gr. 12. Rule. To reduce decimal fractions of higher denominations to whole numbers of lower denominations : Multiply the decimal fraction hy what it takes of the next lower denomination to make a unit of the denomination of the given decimal, pointing off as in multiplicatiori of decimals; so continue till the number is reduced as low as is required. Examples. Reduce to whole numbers of lower denominations, 6. 1.0004^ of a bushel. 1. .8975 of a week. Ans. 6 d. 6 h. 46 m. 48 s 2. 5.624 £. 3. .0074623 tb. 4. .7587565 hhd. 5. .375 of a fathom. ^^ For Dictation Exercises, see Key. 7. .319iofabbL (31 gall.) 8. .578 cord. 9. .0756 of a degree. 10. 2.834 of 1 solid yard. 11. .086 of a Julian year. 164 DECIMAL FRACTIONS. S46. Questions for Review. What are Decimal Fractions? How are they generally written? how read ? How distinguished from whole numbers ? Which figure indicates the denomination? What is the name of the first place at the right of the point? of the second? third? fourth? fifth? sixth ? Which is the place of thousandths? of millionths? of billionths? of trillionths ? Give the rule for reading a decimal fraction. Read 7.05 as a mixed number j as an improper fraction. Read .20 and .21 so that they may be distinguished. Read .504 and 500.004. Is the value of a decimal fraction altered by annexing ciphers? What is changed ? Why does the value remain the same ? What is the effect of placing a cipher between the decimal fraction and the point ? Give the rule for writing decimal fractions. Rule for Addition ; for Subtraction ; for multiplying by 10, 100, 1000, &c. ; for dividing by 10, 100, &c. ; general rule for multiplication. Illustrate the rule by an example, and give the reason for pointing off. Give the rule for division of decimals. Perform an example to illus- trate the rule, and explain. When the dividend does not contain the divisor what must be done ? Rule for reducing common fractions to decimals. Illustrate and explain. Rule for reducing a decimal to a common fraction. Illustrate and explain. What fractions cannot be reduced wholly to the decimal form ? What are they called ? What are the repeating figures called? How is a.repetend distin- guished ? Rule for reducing circulating decimals to common fractions. Rule for reducing a compound number to decimals of higher denom- inations. Illustrate. Rule for reducing decimals to whole numbers of lower denomina- tions. Elustrate. miscellaneous examples. 165 347* Miscellaneous Examples. 1. What is the amount of 3.75 tons, .085 tons, 1.17| ton^ and 1 cwt. 3 qr. ? 2. What will be the interest on $585, for 6 days, if the interest )n $1 be $.001 ? 3. What is the amount of $75823 for 7 y. 3 m. 15 d., if the amount of $1 for the same time be $1.510416§? 4. At $1.33J a pair, how many cases of shoes, of 63 pairs each, can be bought for $936? 5. How many acres of land in a lot which is 105 rd. 4 yd. 1-^ ft. long, and 100.356 rd. wide ? 6. Required the price of three boards at $.03^ per sq. ft., the boards being of the following dimensions : 17.75 ft. by 1 ft. 3 in. ; 15 ft. 10 in. by 1.37^ ft., and 13.5 ft. by .916§ ft. 7. What is the amount due for the following ? 5200 ft. of boards at $20 per M. 700^ ft. " « « 22.50 per M. 94 ft. " " « 36 « 8. If 3 hhd. 42 gall. 2f qt. of molasses cost $92.64, what is the price per hhd. ? 9. What is the cost of board for 7 y. 10 m. 18 d., at $200 per jrear? What cost 10. 9 gall. 3.4 qt. of vinegar, at $.12^ per gall. ? 11. 30 ch. 1 rd. 15 1. of a canal, at $3550 per mile? 12. 5 bu. 2 pk. 3 qt. of wheat, at $1.25 per bu.? 13. 47 gross, 10 doz. pens, at 4 s. 6d. per gross? 14. 6 lb. 7 oz. 6 pwt. 7 gr. of gold, at $16.30 per oz. ? 15. 17f yds. of ribbon, at $.19 per yd.? 16. 12J- doz. chairs, at $1.90 apiece? 17. A road 9 m. 3 fur. 12^ rd. long, at $2475 per mile? 18. 12520 oranges, at $2f per hundred? 19. 3 cwt. 40 lb. herring, at 12 s. 6d. per cwt.? 20. At $4 per bu. how many bu., pk. and qt. can be bought for $l5.37i ? 166 PRACTICE. 21. At 1 s. 9 d. per lb. what cost 3580.5 lb. hides ? 22. If 5 lb. 5 oz. of beef cost $.564||, what is the price per lb.? 23. Required the cost of 19 gall. 3 qt. 1 pt. of oil, at 2 s. 6 d. per gall. 24 If 25375 feet of boards cost $240,555, what is the price per M. ? 25. What is the cost of 7 §, 5 3, 2 B, of medicine, at $.96 per lb.? 26. How many cords in a load of wood, 6.5 ft. long, 4.8 ft. wide, and 3.2^ ft. high ? 27. How many casks gauging 10.485 gall, can be filled from a hogshead gauging 83.88 gall. ? 28! What will be the cost per sq. yd. if $157,675 are paid for laying 4 pieces of sidewalk, measuring as follows : 40§ ft. by 4 ft., 75 ft. by 7.84 ft., 8 ft. 10 in. by 4.5 ft., and 100 ft. by 18.37 J ft.? PRACTICE. 248. Practice is the process of finding the value of a quantity by operating upon an assumed value, or by combining the values of convenient parts. III. Ex., I. What cost 5750 lbs. tea, at 37^ cts. per pound? Operation. 5750 lbs. at $1 per lb. will cost $5750. « « " $.25 " « « ^ of 5750 = 1437.50 « « " .12^ « « " I of 1437.50 z=z 718.75 $2156.25, Ans. OR, 5000 lbs. at .37^ will cost 5000 X .375 =: $1875. 500 « « " " " ^Ig- of 1875 =: 187.50 250 " " « " " 4 of 187.50 — 93.75 5750 lbs. at .37^ per lb., = $2156.25, Ans> PRACTICE. 16? III. Ex., II. What is the price of 17 A. 3 R. 25 rds. of land, at $200 per acre ? Operation. If the price of 1 acre is $200. the price of 17 acres will be 17 X 200 = 3400. « « " 2 roods " " ^ of 200 z=i 100. « « « 1 « " " 1 of 100 =z 50. « " " 20 rods « " ^ of 50 1= 25. " « « 5 " " " ' ^ of 25 z^z 6.25 « « « 17 A. 3 R. 25 rds. — $3581.25, Anjt. III. Ex., III. What cost 5 T. 13 cwt. 1 qr. 10 lbs. of hay, at $16.67 per ton? Opkration. If the price of 1 ton : = $16.67 the price of 5 tons will be 5 X 16.67 . = 83.35 " « " 10 cwt. " " ^ of 16.67 : =1 8.335 « " « 3 " " " ^\ of 8.335 : = 2.500+ « " " 1 qr. " " i of ^ig of 8.335 = .208+ « " " 10 lbs. " " 1 of .208 : =: .083+ « « « « T IS rwf. 1 nr- 10 lbs. -- 1. General Eeview, No, 5, PART I, 1. Multiply 675 by -j^^, the product hy 100; divide the last product by iOOO, this quotient by i,\ multiply ih.Q last quotient by g^gj, and add the five results, 2, From ,1 lb. take .0678 lb. ^. Multiply 8.05^ by .056|, 4. What is the cost of « 4 ft, of hoards, at $20 per M. ? 5. .007644 -r- 36 = ? 6. 8.052-^.0044 = ? 7. 10. -r- 1000 = ? 8. .065455 ~ .065 = ? 9. Reduce ^q to a decimal* 170 GENERAL REVIEW. 10. .3^+.92|r=? 11. What is the cost of 3 pk. 7 qt. of peas, at $3.75 per bushel ? 12. Reduce .8765f deg. to whole numbers of lower denomina- tions. 13. Reduce .015 and .0096 to cpmmon fractions- 14! Reduce .39 and .3432 to common fractions. PART II. 1. "What is the largest number that will exactly divide 475200 and 216000? 2. In 324 sheets of lead, each 12 in. by 8 in., -^ in. thick, how many solid inches ? 3. In 20692 sq. rd. how many acres ? 4. Reduce 19 h. 12 m. to the fraction of a day. 5. What is the 5th power of .09 ? 6. 50.76§ + .834| + .0^r=? 7. If 6 yd. of cloth cost $51 , what will 14f yd. cost ? 8. Add^£, f s., and|d. 9. Reduce .21 pt. to the decimal of a peck. 10. How many inches in length of that which is 8^ in. in • breadth will make a square foot ? 11. Reduce yfg- to a decimal fraction. 12. Carry out the following bill : — Franklin, Dec. 14, 1864. B. Frank Watson, Bought of B. Cooudge; 10850 ft. Boards, at $11. per M., 3000 « « « 19.375 " 2500 « Lathing, « 4.75 « 1500 " Shingles, '^6. « 250 « Plank, « 13. « 1250 " Timber, « 12.80 " 4220 Bricks, " 12.50 " Received payment, ' B. COOLIDGE. t^ For Changes, see Key. PERCENTAGE. 171 PERCENTAGE. 9«>3. The subject of Percentage comprises operations in hundredths. Per cent., from the Latin per, by, and centum, hun- dred, signifies by the hundred ; thus, 4 per cent, of any number c,Y quantity is j^^ of that number or quantity. Any per cent, may be expressed as a decimal fraction, as a common fraction, or by the use of the following sign, % ; thus, 4 per cent is written .04 or T^^ or 4%. H " u " " .03| « tI^ " H%' lA" (C " " .OlA " 4t^t. - 1^V% i " ii " " m " j^ - i%. Examples. Represent the following rates decimally : — 1. 8%. Arts. .08. 4. 13%. 7. 16|%. 2. 20%. Ans. .20. 5. 6i%. 8. 106%. 3. 12^%. 6. i%. 253* III. Ex. Reduce 5 per cent, to its lowest terms. Examples. Reduce the following rates to their lowest terms : — 1. 90%. Ans.j\, 5. 12^%. 2. 50%. Ans, ^. 6. 8^%. 3. 75%. 7. 16§%. 4. 40%. 8. 33^%. 9. 125%. 10. 621%. 11. 87^%. 12. 31^%. 254L, III. Ex. Reduce g to a per cent. (Art. 238.) 8 ) 3.00 .d1^ = SlWc,Ans. 172 PERCENTAGE. Examples. Reduce the following fractions to a per cent. : — 1. fo- Ans.SOfo- 4- 1- 1 7. J. 2. |. Ans. 62J^c- 5. 3^. 8. lA- 3. A-. 6. W. 2S5, III. Ex. Find the complement of 15 per cent., t. e., what it wants of being 100 per cent. 100% — 15% =85%, ^715. Examples Find the complement of the following rates : — 1. 92%,. Ans. 8%. 2. 51%.^?*s. 49%. 3. 11%. 4. 83^%. 7. 18|%. 5. 87^%. 8. 56|%. 6. 33^%. S50. To Find any Per Cent, of a Number. III. Ex. What is 25% of 76 bushels of grain ? 76 bu. X .25 =: 19 bu., Ans. or 25 ^ =1 ; ^ of 76 bu. = 19 bu., Ans. Hence the Rule. To find any per cent, of a number : Multiple/ hy the rate per cent., expressed decimally. Examples. What is 1. 8% of $800? Ans. $64. 2. 5% of 324.40? ^ws. 16.22. 3. 20% of 375 men? Ans. 75 men. 4. 71% of 800 trees? Ans. 60 trees. 5. 12% of 78 bu.? Ans. 9/^ bu. 6. J% of $14.40 ? Ans. $.108. 7. 4^% of 12£. 6s.? Ans. lO^s. 8. 235% of $85? Ans.%\^'dl. 9. 5% off? Ans.^\. 10. 4% of 110% of 8750? Ans. $33, 11. ^% of $200.75? 12. 4^% of 1000 gall.? 13. 10% of49£. 7s. 6d.? 14. 66§% of8d. 5h. 36 m.? 15. I f Lincoln ; the Cr. side what he has sold to Lincoln. PARTIAL PAYMENTS. 187 PARTIAL PAYMENTS. Notes. 368* A Promissory Note, usually called a Nbte^ is a Ivritten promise to pay money or merchandise for value re- ceived. S69. The gum promised is called the Principal or Face of the Note, and should be written in words. 370. In order that a note shall draw interest from date, '^with interest " must be inserted in it ; but notes on demand^ without interest specified, draw interest from the time payment is de- manded ; and notes on time, from the time when due, if not then paid, though interest is not specified. S71, To be negotiable, i. e., transferable or salable, a note must be made payable " to order" or "bearer." If " to order," the holder cannot transfer it without endorsing it, i. e., writing his name on the hack of it. The endorser of a note becomes liable for its payment under certain circumstances. He may endorse, without becoming lia- ble, by writing above his name " without recourse." 272m According to custom, and, in many States, by law, a note is not considered due till three days after the time specified for its payment. These are called days of grace, and interest is taken for them. Note. — A note is said to mature when it becomes due. 273, Partial Payments are payments in part of notes or other obligations. 374, To compute the interest on notes, when partial pay- ments have been made, observe the following, called The United States Rule. 1. Find the amount of the sum due from the time interest com- mences to the time of the first payment ; subtract the payment, if it will cancel the interest, and consider the remainder a new prin- cipal ; find the amount of this new principal from the time of ths 188 PERCENTAGE. Jlrst payment to the time of the second ; subtract the second pap'^ ment as before, and so proceed to the time of settling the note. 2. By the decisions of the United States Court, when a pay- ment will not cancel the interest due, interest is computed on the principal till surfficient sums have been paid to cancel all the in^ terest due, when all the payments are subtracted from the amount due as if paid in one sum. Note. — When partial payments are made, the account is kept by en- tering the same, with their dates, on the back of the note. The entrioi are called endorsements. Examples. 1. Suppose a note for $1908.42, dated Aug. 9, 1851^10 be on interest till Feb. 15, 1852, when a payment of $1732.59 is made; what sum will remain due ? Ans. $234,991. 2. Suppose the above balance ($234,991) to remain on inter- est till April 3, 1853, when another payment of $50 is made; what will then be due ? Ans. $200.97. 3. Suppose the balance ($200.97) to continue on interest to Jan. 9, 1860, what will be due at that time ? * 4. A note for $75.83, with interest, is dated Jan. 19, 1850; suppose $15 to be paid July 15, 1850, what will remain due ? 5. Suppose $40 of the above balance to be paid April 13, 1852, what will then be due ? 6. If this balance remains on interest till Feb. 7, 1853, what sum will then be due? Ans. $31,105+. 7. A note for $50, dated Jan. 1, 1862, is on interest at 6% till April 15, 1862, when a payment of $25.87 is made. What sum will remain due ? 8. If the balance of the above should remain on interest at 7% till Jan. 13, 1863, what will then have to be paid to discharge the note? Promissory Note. (Art. 268.) 9. $300. Philadelphia, April 5, 1857. On demand, I promise to pay E. Varnum, or bearer, three hundred dollars, with interest, value received. C. J. Potter, PARTIAL PAYMENTS. 189 On the above note were the following endorsements : — Eeceived, May 29, 1860, $217.49. " Apr. 23, 1862, $50. What will be the balance due on the above note December 15, 1869? Am. $153,275. Operation. Principal, $300. , on interest from Apr. 5, '51. Interest on principal, . 56.70 , to May 29, '60, (3 y. 1 m. 24 d.) Amount, $356.70 let payment, .... 217.49 2d principal, .... $139.21 , on interest from May 29, '60. Interest on 2d principal, 15.869, to Apr. 23, '62, (1 y. 10 m. 24 d.) Amount, $155,079 2d payment, .... 50. 3d principal, .... $105,079, on interest from Apr. 23, '62. Interest on 3d principal, 48.196 , to Dec. 15, '69, (7 y. 7 m. 22 d.) Amount, $153,275,^/15. 10. $1000. Burlington, Oct. 5, 1854. For value received, I promise to pay to the order of Joseph P. Battles, one thousand dollars, with interest, on demand. J. BUSNELL. ENDORSEMENTS. Received of within Dec. 8, 1854, $125. « « " May 12, 1855, 316. « <' " Sept. 2, 1855, 417. « " " Mar. 9, 1856, 100. What balance remained due June 15, 1856? Ans. $93.353 -f . 11. $700. Lancaster, April 5, 1847. On demand, with interest at 7%, we promise to pay H. K. Oliver, or order, seven hundred dollars, value received. Warren Burton & Co. ENDORSEMENTS. Received of within, Oct. 29, 1850, $217.49. « « July 23, 1852, 200.00. What remained due Dec. 12, 1859 ? ^/i*. $814,681. 190 PERCENTAGE. Note. — In the following examples in Partial Payments, consider each note to be on demand with interest from its date, unless otherwise specified. 12. A note for $960 is dated Nov. 16, 1855, on which was paid $140, Nov. 11, 1856; $80, July 30, 1857; $70, Jan. 2, 1858 ; and $100, Dec. 1, 1858. What balance is due Oct. 30, 1859, reckoning interest at 7% ? Ans. $806,077. 13. $350. Bristol, Jj!?rz7 5, 1850. For value received, we jointly and severally promise John Ingalls to pay him, or order, three hundred fifty dollars, on demand, with interest at 5 ^ per annum, after three months from date. Hood & Bishop. On this note were the following endorsements: Nov. 1, 1852, received $87 ; March 7, 1855, received $150 ; Feb. 19, 1858, received $115. What was due Sept. 15, 1862 ? Ans. $125.Gl-f . 14. A note for $935 is dated Sept. 1, 1855, on which was paid $125.75, Jan. 15, 1856 ; $250, March 25, 1861 ; $300, IMay 10, 1861. What was the balance due July 1, 1861 ? Ans. 1549.713. Opekation. Principal, $935. , on interest from Sept. 1, 1855. Interest on principal, . 20.881, to Jan. 15, '56 (4 m. 14 d.). A.mount, $955,881 1st payment, .... 125.75 2d principal, .... $830,131, on interest from Jan. 15, ^5Q. Interest on 2d principal, 258.724, to Mar. 25, '61 (5 y. 2 m. 10 d.). Interest on 2d principal, 6.225 , to May 10, '61 (1 m. 15 d.*). Amount $1095.08 2d and 3d payments, . 550.00 , both required to cancel interest. 3d principal, .... 545.08 , on interest from May 10, '61. Interest on 3d principal, 4.633, to July 1, '61 (1 m. 21 d.). Amount, $549,713, Ans. 15. $500. Salem, April 1, 1855. For value received, I promise to pay W. J. Rolfe, or order, five hundred dollars, on demand, with interest from Oct. 1, 1855. Irenas Edwards. ♦ See Art. 274, Rule, 2d Clause. PARTIAL PAYMENTS. 191 ENDORSEMENTS. Received of the within, Apr. 1, 1856, $12. " " " " Apr. 1, 1857, 1100. « " « " Apr. 1, 1858, 1100. What is due June 19, 1858? Ans. $363,646+. 16. A note for |1000 is dated June 1, 1860. The endorse- ments are : $75, paid Aug. 1, 1860 ; $125.75, paid Dec. 15, 1860 ; $250, paid Feb. 25, 1866; and $300, paid Apr. 10, 1866. What will be due on this note June 1, 1866? Ans. $549.713-f . 17. A note for $790, dated Oct. 9, 1862, is endorsed Sept. 6, 1863, with $320; Jan. 30, 1864, with $10; Oct. 9, 1864, with $190. What balance is due Feb. 3, 1865, interest at 5% ? Ans. $338.77+. iSr A note for $800, dated Jan. 15, 1860, is on interest after 6 months, and is endorsed Apr. 18, 1861, $100; Jan. 1, 1863, $70; and June 15, 1864, $62.50, — interest being at 7% . What was due July 15, 1865? Ans. $830,415+ 19! Upon a note of $425, dated July 13, 1859, there are the following endorsements: August 10, 1861, $50; November 18, 1862, $150. What will be due, if the note is settled July 13, 1863? 20! A note for $250, dated May 15, 1838, is endorsed Feb. 25, 1841, $111.66§; Oct. 19, 1842, $15; May 9, 1848, $62; and Oct. 15, 1849, $100.30. Required the balance due July 22, 1851. 21*. A note for $489 is dated Jan. 20, 1850, and endorsed as follows: June 26, 1850, received $50; Feb. 26, 1852, received $40; July 8, 1855, received $90; Jan. 26, 1856, received $200; June 20, 1856, received $200. If this note was on interest from three months after date, what was due Nov. 20, 1856? 275* The following rule for Partial Payments is in general use, when the whole period of time is less than one year : — Rule. Fi7id the amount of the principal for the tvhole time the note is on interest ; find, also, the amount of each payment from the time it is rnade to the time of settling the note ; and de~ duct the sum of the payments^ with their interest, from the amount of the principal. 192 PERCENTAGE. III. Ex. 1800. Burlington, July 7, 1860. Three months after date, I promise to pay John Thetford, or bearer, eight hundred dollars, with interest. Benjamin Stokes. On the back of the above note were recorded the following payments : — Received, Aug. 16, 1860, $200. " Oct. 8, 1860, $480. « Feb. 20, 1861, $49.92. What balance was due at the time of settlement, July 1, 1861 ? Ans. $84.65. Operation. Principal, $800, from July 7, '60, to July 1, '61 (11 m. 24 d.), amounts to $847.20 1st pay't, $200, from Aug. 16, '60, to July 1, '61 (10 m. 15 d.), amounts to . . . $210.50 2d pay't, $480, from Oct. 8, '60, to July 1, '61 (8 m. 23 d.), amounts to . . . 501.04 3d pay't, $49.%2, from Feb. 20, '61, to July 1, '61 (4 m. 11 d.), amounts to ... 51.01 — 762.55 Balance due, . . . Ans. $84.65 1. $10000tV^. Concord, OcL 4, 1863. In two months from date, I promise to pay to the order of Benjamin Tyler, at Suffolk Bank, Boston, ten thousand -^^^^ dollars, with interest, value received. Thomas Beeman. endorsements. Received of within, $672.41, Nov. 5, 1863. « " " $7682.42, Nov. 15, 1863. " ** « $437.98, Nov. 16, 1863. " " " $833.42, Nov. 19, 1863. What was the balance due on the above note, when the note became due ? Am, $443,555. PARTIAL PAYMENTS. 193 2. $1200 Albany, April 1, 18G2. One year from date, for value received, I promise to pay to J. V. Smiley or order, twelve hundred dollars, with interest, at 7^. Okrin Jones. The above was indorsed as follows : — April 11, 1862, $161.08; July 18, 1862, $224.14; July 27, 1862, $17.90; Jan, 28, 1863, $100,25. What was still due April 1, 1863? Ans, $756,565. Q76. The following is the Connecticut Rule. 1. When a year's interest or more has accrued at the time of a payment, and always in case of the last payment, foUow the Governmeivt Eule, (Art, 274.) 2. When less than a year's interest has accrued at the time of a payment, except it be the last payment, find the difference between the amount of the principal for ■an entire year, and the amount of the payment for the balance of a year after it is made ; this difference will form the new prin-cipal. 3. If the interest which has arisen at the time of a payment exceeds the payment, no interest will be 'Computed upon the pay- ment, but only up&n the principal. 1, $1000, Hartford, March 9, 1855. In one year from date, for value received, I promise to pay Geo. Yates or order, one thousand dollars, with interest, at 6^. Joseph W, Boomer, Jr. ENDORSEMENTS, Received Nov. 19, 1855, $204; Mar. 3, 1857, |50; June 15, 1858^ $600 ; Nov. 1, 1858, $85. What balance was due Jan. 1, 18^9 ? Ans. $241,798. 13 194 PERCENTAGE Operation. $1060 Amount of principal from Mar. 9, '55 to Mar. 9, '56, (1 yr.). 207.74 " 1st payment from Nov. 19, '55 to " " (3fm.). 852.26 Balance, forming 2d principal. 51.135 Interest from Mar. 9, '56 to Mar. 9, '57, (1 yr.). 903.395 Amount. 50. 2d payment, being less than interest, has no interest. 853.395 Balance, forming 3d principal. 64.857 Interest from Mar. 9, '57 to June 15, '58, (1 yr. 3 m. 6 d.). 918.252 Amount. 600. 3d payment, time being more than 1 year, has no interest, 318.252 Balance, forming 3d principal. 10.396 Interest from June 15, '58 to Jan. 1, '59, (6 m. 16 d.). 328.648 Amount of 3d principal to time of last payment. 85.85 Amount of $85 from Nov. 1, '58 to Jan. 1, '59, (2 m.). $242,798, Ans. Balance due Jan. 1, '59. S77. Annual Interest. III. Ex. What is the amount due on a note for $1000, inter- est payable annually, if no payment should be made till the ex- piration of 4 y. 6 m. 12 d. ? The holder of this note should be allowed interest on the interest from the time it is payable to the time of settlement, in addition to the interest upon the note. The int. on $1000 for 4 y. 6 m. 12 d. = $272.00 " " $60 for 3 y. 6 m. 12 d.^ 2 y. 6 m. 12 d. $60 " 6 m. 12 d. = the int. on $60 ly. 6 m. 12d. Kor 8 y. 1 m. 18 d > — 29.28 Principal, 1000.00 Amount due, $1301.28, Ans. The interest is first taken upon the face of the note for the full tinic • then upon the $60 due at the end of the first year for the balance of the time for which the note has to run ; and so on for the other payments. Hence the Rule for Annual Interest. Compute interest on the principal for the entire time it is on interest ; compute interest COMPOUND INTEREST. 195 also upon one year's interest for the sum of all the periods of time for which each yearly interest remains unpaids The sum of the interests thus found will he the annual interest. Examples. 1. What is the annual interest of $200 for 4 y. 6 m. 3 d.? Ans. $59,884. 2. What is the annual interest of $334 for 3 y. 8 m. 10 d. ? Ans. $80,148+. 3. What is the annual interest of $118.50 for 5 y. 3 m. 18 d. ? Ans. $42,588+. . 4. What is the amount at annual interest of $175 for 6 y. 2 m. 25 d.? ^ws. $250,821+. Note. — For New Hampshire rule for annual interest with partial pay- ments, see Appendix, page 334. COMPOUND INTEREST. S78, Compound Interest is interest on both principal and interest, the sum of the two forming a new principal at specified intervals of time. Note. — Interest may be compounded, or added to the principal, annu- ally, semi-annually, or for any period of time agreed upon. 379. III. Ex., I. What is the compound interest of $212 for 2 y. 5 mo. 6 d., at 6% ? Operation. Principal, Amount of $1 for 1 year, $212 1.06 « $212 for 1 year, 224.72 1.06 « $212 « 2 years, " $1 « 5m.6d., 238.2032 1.026 " $212 " 2y. 5 m. 6d., Principal, subtracted, 244.3964832 212. Compound Interest, $32,396+, Ans. 196 PERCENTAGE. EuLE FOR Calculating Compound Interest. Find the amount of the principal to the time when interest is first due ; find the amount of this sum for a second period of time as at first, and so on till the entire periods of time for which interest is to he compounded are exhausted; find the amount for the balance of time as in simple interest. This will he the amount at compound interest. To obtain the compound interest, subtract the first principal. Examples. 1. What is the amount of $350 for 3 years, at 6% ? Ans. $416,855+. 2. What is the compound interest of $250.50 for 4 years, at ^% ? Ans. $53,984+. 3. What is the amount of $1000 for 3 y. 11 mo., at 6% ? Ans. $1256.521+. 4. What is the compound interest of $427.56 for 3 y. 7 m. 6 d., at Ans. $100,003. 5. What is the compound interest of $350.60 for 2 y. 11 m., at 7' Ans. ).558+. 6. What is the amount of $250 for 1 y. 3 m. 18 d., at 5% per annum, interest compounding semi-annually ? . Operation. Amount of $1 for 6 mo., $1,025 Multiplied by 250 Amount of $250 for 1st 6 mo., 256.25 Multiplied by amount of $1 for 6 mo., .... 1.025 Amount of $250 for 1st 12 mo., 262.656+ . Multiplied by amount of $1 for 3 m. 18 d., . . . 1.015 Amount of $250 for 1 y. 3 m. 18 d., . . . Ans. $266,595+ 7. What is the amount of $30 for 1 y. 2 m., at 6% per an- num, interest compounding semi-annually? Ans. $32,145+. 8. What is the compound interest of $800 for 1 y. 1 m., at 6^, interest compounding quarterly? Ans. $53,336+. 9. What is the compound interest of $240 for 8 m. 15 d., at 10% per annum, interest payable semi-annually? 10. Find the compound interest of $80 from Sept. 1 1860 to Oct. 7, 1861, at 8 %. COMPOUND INTEREST. 197 11. What is the amount of $1400 for 1 y. 2 m. 28 d., interest compounding at the expiration of every 4 months ? 12. Required the amount of $700 from Jan. 5, 1864, to Nov. 21, 1865, at 4% per annum, interest payable semi-annually. 380, The process of computing compound interest may be shortened by the following Table, Showing the amount of $1 or £1 at compound interest from 1 year to 30 years, at 3, 4, 4^, 5, 6, and 7^. Year. 1 3 p. cent. 1.030000 4 p. cent. 4| p. cent. 5 p. cent. 6 p. cent. 7 p. cent. 1.040000 1.045000 1.050000 1.060000 1.070000 2 1.060900 1.081600 1.092025 1.102500 1.123600 1.144900 3 1.092727 1.124864 1.141166 1.157625 1.191016 1.225043 4 1.125509 1.169859 1.192519 1.215506 1.262477 1.310796 5 1.159274 1.216653 1.246182 1.276282 1.338226 1.402552 6 1.194052 1.265319 1.302260 1.340096 1.418519 1.500730 7 1.229874 1.315932 1.360862 1.407100 1.503630 1.605781 8 1.266770 1.368569 1.422101 1.477455 1.593848 1.718186 9 1.304773 1.423312 1.486095 1.551328 1.689479 1.838459 10 1.343916 1.480244 1.552969 1.628895 1.790848 1.967151 11 1.384234 1.539454 1.622853 1.710339 1.898299 2.104852 12 1.425761 1.601032 1.695881 1.795856 2.012197 2.252192 13 1.468534 1.665073 1.772196 1.885649 2.132928 2.409845 14 1.512590 1.731676 1.851945 1.979932 2.260904 2.578534 15 1.557967 1.800943 1.935282 2.078928 2.396558 2.759031 16 1.604706 1.872981 2.022370 2.182875 2.540352 2.952164 17 1.652848 1.947900 2.113377 2.292018 2.692773 3.158815 18 1.702433 2.025816 2.208479 2.406619 2.854339 3.379932 19 1.753506 2.106849 2.307860 2.526950 3.025599 3.616527 20 1.806111 2.191123 2.411714 2.653298 3.207135 3.869684 21 1.860295 2.278768 2.520241 2.785963 3.399564 4.140562 22 1.916103 2.369919 2.633652 2.925261 3.603537 4.430401 23 1.973587 2.464715 2.752166 3.071524 3.819750 4.740529 24 2.032794 2.563304 2.876014 3.225100 4.048935 5.072366 25 2.093778 2.665836 3.005434 3.386355 4.291871 5.427432 26 2.156591 2.772470 3.140679 3.555673 4.549383 5.807352 27 2.221289 2.883369 3.282009 3.733456 4.822346 6.213867 28 2.287928 2.998703 3.429700 3.920129 5.111687 6.648838 29 2.356565 3.118651 3.584036 4.116136 5.418388 7.114256 30 2.427262 3.243397 3.745318 4.321942 5.743491 7.612254 198 PERCENTAGE. III. Ex., II. What is the compound interest of 1520, at 7%, for 4 y. 1 m. 24 d. ? Operation. Amount of $1 at 7^ for 4 y. by the table, . . $1.310796 Multiplied by the principal, 520 Amount of $520 for 4 yrs., 681.613-f- MultipHed by am't of $1 for 1 m. 24 d., . . . . 1.0105 Amount of $520 for 4 y. 1 m. 24 d., . .... 688.769-4- Principal subtracted, 520. Compound interest, . Ans. $168,769-}- 13. What is the conapound interest of $480 for 7 y. 10 m. ? Ans. $277.8294-. 14. What is the amount of $100 for 2 y. 4 m., at 7% ? 15. What is the compound interest of $200 for 3 y. 2 m. 6 d. ? 16. What is the amount of $221,075 for 3 y. 5 m., at 7% ? 17. What is the amount of $280 for 1 y. 10 m. 22 d., interest payable semi-annually ? 18! What is the amount of $50 for 3 y. 13 d., at 5% ? 19! What is the compound interest of $896 for 2 y. 6 m. 15 d., at 5% ? 20! Find the compound interest of $300 for 3 y. 4 m. 12 d., at 21*. Find the amount of £58 for 3 y. 5 m. Ans. 70 £. 16 s. 1+d. 22! Find the compound interest of 75 £. 9 s. 9 d. for 4 y. 8 m; 27 d., at 5%. 23! What is the difference between the compound and simple interest of $678.25 for 3 y. 6 m. 6 d. ? Ans, $11,488—. 24! What is the difference between the compound and simple interest of $100 for 1 y. 4 m., the compound interest payable semi-annually ? 25! What is the difference between the amount of $175.08, at compound and at simple interest, from May 7, 1861 to Sept. 25, 1863, at 7% ? PROBLEMS IN INTEREST. 199 26* Find the difference between the simple and compound in- terest of 04 £. 12 s. 6 d. for 2 y, 6 m. 12 d., at 8% ? l^ For Dictation Exercises, sec Key. PROBLEMS IN INTEREST. 381. Since interest is always the product of the three factors, principal^ rate, and time^ it follows that to find the time, rate, or principal, when the interest and two of the other terms are given, it is only necessary to divide the interest by the prod- uct of the two given terms. 383. To FIND THE Time, when the Interest, Pkin- ciPAL, AND Rate are Given. III. Ex. In what time will f 300 gain |63 interest at 6% ? Operatiox. The interest of $300 for 1 year at Int. of ^300 for 1 y. =$18 e^ j^ $18; it will require as many 1^ ) 6^'^ years for $300 to gain $63 as $18 is T~7 - contained times in $63, which is 3i 3.0 yrs. Ans. . . „, i., ' , ^ times. Ans. 3| y. Hence the Rule. To find the time, when the interest, principal, and rate are given : Divide the given interest by the interest of the principal at the given rate for 1 year. Examples. What time will be required 1. For $400 to gain $20, at 6% ? -^ns. 10 m. 2. For $500 to gain |?60, at 4% ? Ans, 3 y. 3. For 168.25 to gain 1^3.003, at 6% ? 4. For #640 to gain $07.20, at 7% ? 5. For $3000 to gain $205, at 5% ? G- For $408 to gain $170, at 7|% ? * 7. For 1450 to gain $192.30, at 8%? 8. For $280 to amount to $301, at 5% ? Note. — Subtract $280 from $301 to find the interest. 9. In what -time will $200 amount to $400, at 6% ? 10. In what time will $500 amount to $658.33}, at 6% ? 200 PERCENTAGE. 383* To FIND THE Kate, when the Intekest, Time, AND Principal, are known. III. Ex. At what rate per cent, will $250 gain $25 in 2 years ? OPERATiojf. $ If the interest of Int. of $250 for 2 y. at 1^, . . $5 ) 25 $250 for 2 y. at 1 5 per cent, is $5,it will require as many times 1^ to gain $25 as $5 is contained times in $25, which is 5 times. Ans. 5^. Hence the Rule. To find the rate, when the interest,, time, and principal are given : Divide the given interest by the interest of the princi-- pal for the given time at 1 per eent^ Examples. At what % will 1. $360 gain $40.80 in 1 y. 5 m. ? Ans. 8^^. 2. $100 gain $33i in 12 y. 6 m. ? Ans. 2f %, 3. $250 gain $3.75 in 4 m. ? 4. $25 gain $7.87^ in. 3 y. 6 m.? 5. $100 gain $25in7|y.? 6. $48.24 gain $8J1 in 2 y. 9 m. 10 d.? 7. $75 amount to $78.75 in 2 y. 6 m. ? Note. — ^TS-TS — $75 = $3.75 interest. 8. At what rate will $50 amount to $55.25 in 2 y.? 9. At what rate wiU $1000 amount to $1058.334 in 10 m. ? 384* To riNi> THE Principal, when the Interest, Time, ani> Rate are known. III. Exi What principal will yield $42.50 interest in 8. m. 15 d. at 6% ? OpEEuVTIOK-. .f Int. of $1 for 8 m. 15 d. at Q%, . . . $.0425 ) 42.50 (1000. The interest of $1 for 8 m. 15 d. at 6^ is $.0425 ; it will require aa many dollars of principal to gain $42.50 as $.0425 is contained times, in $42.50, which is 1000 times. Ans. $1000. Hence the PROBLEMS IN INTEREST- 201 Rule. To find the j^rincipal, when the interest, time, and rate are known : Divide the given interest hj the interest of 1 dollar at the given rate for the given time. Examples. What principal will gain 1. 115 in 2 y. at 6% ? Ans. |125. 2. $20 in 4 y. at 5% ? Ans. $100. 3. $76.50 in 2 y. 6 m. at 3% ? 4. $1,705 in 7 ra. 15 d. at 4% ? 5. $68,990 in 1 y. 4 m. 24 d. at 5% ?. 6. $4,128 in 11 m. 14 d. at 6% .^ Note. — (4.128 -^ .057 1). Reduce dividend and divisor to thirds before dividing. Ans. $72. 7. What principal will be required to gain $24 in 60 days at 2% a month? 8. What principal must be on interest 2 y. 5 m. 29 d. at 6 % to gain $89.40 ? 9. What is the principal which being on interest at 7% per annum, gains $62.50 semi-annually.'^ S85, To FIND THE Principal, when the Amount, Time, AND Rate are known. III. Ex. What principal will amount to $17,238 in 2 m. 12 d. at 7% ? Operation. The amount of 1 dollar for 2 m. 12 d. is $17.238 -f- $1,014 = 17. $1,014; it will require as many dollars to amount to $17,238 as $1,014 is contained times in $17,238, which is 17 times. Ans. $17. Hence the Rule. To find the principal, when the amount, time, and rate are known : Divide the given amount hy the amount of 1 dollar at the given rate for the given time. Examples. What principal will amount 1. To $870 in 7 y. 6 m. at 6% ? Ans. $600. 2. To $537.50 in 2 y. 6 m. at 6% ? Ans. $467.39/^. 202 PERCENTAGE. 8. To $2072.25 in 30 da. at 5% ? * Ans. $2063.651-|-. 4. To $412 in 90 da. at 1% a month? A7is. $400. 5. To $100 in 3 y. 6 m. at 5i% ? 6. To $343.75 in 2 y. 1 m. at 7% ? 7. To $206.25 in 7 m. 15 da. at 5% ? I^^ For Dictation Exercises, see Key. PRESENT WORTH AND DISCOUNT. 386, This subject is a practical application of Art. 285. It embraces all examples in which it is required to know what sum will equitably discharge a note or debt at a given time before it is due. SST, The Present Worth of any sum of money due at a future time without interest, is such a sum as put at interest at the given rate w^ill amount to the debt when it becomes due. It is evident that where money is worth 6% a year, $106 due in one year is the same in value as $100 paid now ; for $100 put at interest for 1 year will amount to $106. S88, Discount is that part of an obligation which is abated or given up when the payment is made before it becomes due, and should in justice equal the interest upon the present worth for the given time. 389, III. Ex. What is the present worth of $210 due 1 year hence, money being worth 5 % ? Here $210 is the amount of some principal for 1 year at 5^; $1^ amounts to $1.05 in a year ; hence it will require as many dollars to amount to $210 as $1.05 is contained times in $210, which is 200 times. Ans. $200. Hence the Rule. To iind the present worth : Divide the given sum by the amount of 1 dollar at the given rate for the given time. The discount of the above ($210) is found by subtracting $200 from $210 ; this leaves $10, which is precisely the same as the interest of $200 for 1 year at 5^. Hence the KuLE. To find the discount : Subtract the present worth from the given sum. PRESENT WORTH AND DISCOUNT. 203 Examples. 1. Find the present worth of 827,50 for 1 y. 8 m. at 6%. 27.50-^1.10 = 25. Ans. $25. 2. Fmd the present worth of $100.90 for 8 mo. at 6%. Ans. $97,076 -f-. 3. Find the present worth and discount of $200 due in 3 mo. at G%, Ajis. $197.0444- pres, worth; $2.956 — discount. 4. What is the discount of $100 for 9 mo. at 4% .^ A?is, $2.91 2-f-. 5. What is the present worth of $1609.30 for 10 m. 24 d. at 5%? Ans. $154.0. 6. What is the present worth of $175.80 for 9 m. 20 d. at 6% ? 175.80 -^ 1.048^ =: 527.40 ^ 3.145 = 167.694-f-. Ans. $lG7.G94-(-. 7. What is the discount of $661,375 for 3 m. 15 d. at G% ? Ans. $11,375. 8. What is the present worth of $96,347 for 8 m. 3 d. at 7% ? A71S. $92. 9. Find the present worth and discount of $75.50 for 8 m. 10 d. Ans. 172.48 pr. w. ; $3.02 disc. 10. Find the present worth of $800.75 for 1 y. 1 m. 10 d. Ans. $750,703+. 11. What is the present worth of $75.85 due in 4 m. at 5% ? 12. What is the present worth of $221,075 due in 3 y. 5 m. at 7% ? 13. If a note for $500 be due in 2 years without interest, what is its value at the present time, money being worth 7% ? 14. What is the present value of $50 due in 3 y. 13 d., inter- est being 5%? 15. A note for $240 is dated June 1, 1860, due in 8 m. 15 d. ; what money will discharge it at date ? Ans. $230,215-}-. 16. A note for $500 is dated April 6, due in 90 days ; what money will discharge it at date? 17. What would discharge the above June 23 ? Ans. $499+. 18. A note for $2000, dated July 15, was given for 1 year, without interest ; what will discharge it at date ? 204 PERCENTAGE. 19. What will discharge it Oct. 15 of the same year? ^m. $1913.875-f. 20.* What sum paid down will discharge a, note of $500, due in 2^ years, the rate being 5% ? 2 it What is the cash value of a note for ^927.60 on 7 days' credit ? 22! What is the value of a note for $139.50 Dec. 11, 1863, which is dated Sept. 9, 1863, and given for 1 year ? $251.90. Trenton, April 1, 1862. In nine months from date, I promise to pay J. Adams, or bearer, two hundred fifty-one j®^^ dollars, value received. C. Quint. 23! What will discharge the above at its date, the rate of dis- count being 6% ? 24? What will discharge the above April 16, 1862, the rate being 7% ? j^^ For Dictation Exercises, see Key. BANK DISCOUNT. 390. Bank Discount is an allowance made to a bank for advancing money on a note before it is due. ^91. Bank discount is the interest on the face of the not© or its amount at maturity for the time it is discounted (Art. 272). 39^, The holder of a lote discounted at a bank receives the face of the note minus the discount. This is called the present worth, the proceeds, or avails of the note. III. Ex. What is the bank discount on a note of $300 for ? What are the avails ? Operatiox. The interest on $300 for 4 mo. is $6.00 « « « $300 for 3 d. is .15 " $300 for 4 m. 3 d. is $6.15, discount. $300 —$6.15 = $293.85, avails of note. BANK DISCOUNT. 205 Hence the Rule. To find the Bank Discount : Compute simple interest on the given sum fourth e time it is to remain on interest, plus three days of grace. To find the avails : Subtract the discount from the given sum. Note. Suppose the above (111. Ex.) was a 6 months note dated Jan. 5, which was to be discounted at a bank March 5, the operation would be precisely the same; the note would mature f July 5, with grace, July 8, and would be discounted for the time to elapse between March 5 and July 8, which is 4 months and 3 days. Examples. Find the bank discount 1. On $75 for 30 days. Ans. I.412+. 2. On a 90 days note for $500, dated May 10, and discounted June 9. Ans. $5.166-|-. 3. On a 60 days note for $256.84, dated Oct. 28, and dis- counted Nov. 12. Ans. $2,054-]-. 4. On $1000 for 3 mo. at 7%. Ans. $18,083+. 5. What are the avails of a note of $700, discounted at a bank for 69 days ? Ans. $691.60. 6.* A trader buys 900 pairs of shoes at $.75 a pair cash, and immediately sells them at $.90 on a note payable in 4 months without interest ; suppose he gets his note discounted at a bank for the 4 months, what will he have made ? Ans. $118,395. ^^ For Dictation Exercises, see Key. 39*S« To FIND FOR WHAT A NOTE MUST BE GIVEN, WHICH^ DISCOUNTED AT A BaNK, WILL YIELD A CERTAIN SuM. III. Ex. What must be the face of that note which, being discounted at a bank for 60 days, will yield $148,425? The bank discount of $1 for 63 d. =r$.0105; $1 -^ $.0105 p- $.9895, avails of $1. $148,425 ~ $.9895 = $160, face of note. If $1 were discounted at a bank, it would yield $.9895 ; to yield $148.42, the note must be given for as many dollars as $.9895 is con- tained times in $148,425, which is 150 times. Ans. $150, t See Art. 272, also Appendix, p. 330. 206 PERCENTAGE. Hence the Rule. To find the face of a note which, discounted at a bank^ will yield a certain sum : Divide the required sum hy $1, minus the hank discount of %1 for the given time at the given rate, and the quotient will he the face of the note. Examples. 1. "What must be the face of a note that it may yield $80 when discounted at a bank for 30 days ? Ans. $80.442~(-. 2. For what must a note on 4 months, without interest, be given, that, when discounted at a bank, it may yield $489.75 ? Ans. S500. 3. For what must a note be given, which is to run 90 days, that- it may yield $400? 4. What must be the face of a note having 60 days to run, that it may yield $989.50 ? 5. For what must a note, dated Sept. 1, on 4 months, be given to yield at its date $400, when interest is 7 % ? 6. For what must a note, dated Jan. 1, payable in 3 months, discount being 5^%, be given, to yield |150? Ans. 1152.162—. 7* For what must a note on 6 months be written, to yield $495.85, when the discount is 7^% ? 11^° For Dictation Exercises, see Key. 3d4:* Miscellaneous Examples in Banking, &c. Note. — All examples in Present Worth or Discount should be con- jsidered in True Present Worth or Discount (page 202), unless Bank Pres- ' ent Worth or Bank Discount is definitely stated. 1. A note for $500, dated July 1, is given for 20 days without interest ; what is its true value July 15 ? Ans. $499.50-f-. 2. What will discharge the above, Aug. 14 ? (Exact days.) Ans. $502. $200. Boston, April 1, 1862. Four months from date, I promise to pay John Bills, or order two hundred dollars, value received. John Orne, Jr. BANK DISCOUNT. 207 3. Suppose the above to be a good and true note, what is it really worth to the holder in cash at its date, money being 6% ? Ans. $196,078+. 4. What could he get from a bank ^r it at its date ? Ans. $195.90. 5. What would he get for it May 1, by true discount? (Art. 287.) Ans. $197,044+. 6. What ought he to get for it April 1, 1863 ? Ans. $208. 7. What is the difference between the avails of a note for $200, payable without interest in 18 months, whether it be paid by true or by bank discount? Ans. $1,586+. 8. What will be the difference between the true and the bank discount of a note for $90.50, due Feb. 9, 1862, and discounted June 15, 1861? Ans. $.177+. $300. Hartford, Juhj 15, 1860. For value received, I promise to pay to the order of myself three hundred dollars, in one year, with interest after six months. John A. Andrew. 9. What sum will the holder of the above receive, if it be discounted at a bank Sept. 15, 1860 ? 10. What sum would the holder of the above receive at its date by true discount ? 11. What would discharge the above note May 8, 1861 ? 12. What would be the bank discount of the above at its date? $500. New Bedford, Oct. 5, 1860. For value received, I promise to pay Alvin Dow, or order, five hundred dollars in three months. Allen Jones. 13. What cash must be paid to discharge the above note at its date by true present worth ? 14. What would be the avails of it at a bank Dec. 5, 1860 ? 15. What would be its real cash value March 17, 1861 ? 16. What would be the true discount on it Nov. 5, 1860? 17. What would be the bank discount of it Nov. 5, 1860? 208 PERCENTAGE. 18* What would be due on the above note Feb. 15, 1862, if $50 had been paid on it at the termination of each six months from its date, interest being 5% ? ^9^. General Review, No. G. 1. Reduce 75%, 16|%, 37|-%, 95%, and 83^% to their low- est terms, and give their sum in a common fraction. 2. If you buy socks at 14.80 per dozen pairs, and sell at $.50 per pair, what % do you gain ? 3. For what must apples which cost $1.25 per bbl. be sold to gain 20% ? 4. If 25% is lost by selling a pair of boots at $4^, what was the cost ? 5. What is the simple interest of $300 from May 5, 1860, to Feb. 2, 1862, at 1^%, a month? 6. What is the amount at compound interest of $271.36 for 2y. 6 m. at 6%? 7. What is the present worth of $4508.25 for 11 days, at 6%? 8. What is the bank discount of $450 for 30 days, at 5% ? 9. What are the avails of a note of $100 discounted at a bank for 27 days? 1 0. What is the amount at simple interest of 5£. 4s. 6d. for 2 years, at 5 % ? 11. The interest of $400 for a certain time at 6% was $60 ; what was the time ? 12. What principal at 5% will gain $4.50 in 10 months? 13. At what Gjo will $462 gain $103.95 in 2 y. 3 m.? 14. For what must a note be given, which, discounted at a bank at 6 % for 60 days, will yield $1295? 15. Given a note for $2500, dated Sept. 5, 1862, on which were paid $50 Jan. 29, 1863, $500 July 1, 1864. The note being on interest at 6 % from its date, what was due Sept. 5, 1864? l^ For changes, see Key. COMMISSION, LIIOKERAGE, AND STOCKS. 209 COMMISSION, BROKERAGE, AND STOCKS. 396. Commissioii is a certain percentage received by a commission merchant for transacting business as factor, or agent, for another. SOT* Brokerage is the percentage received by a broker* A Broker is one who exchanges money and deals in stocks and bills of credit. 398. Stocks are Government Bonds of all kinds, and shares of the capital invested in Banks, Insurance Companies, &c. 399. When stocks and money sell for their original or nom- inal value, they are said to be at par ; when they sell for more than their nominal value, they are said to be at an advance^ above par^ or at a premium ; when they sell for less than their nominal value, they are said to be at a discount^ or helow par, 300. In Commission, the % is estimated upon the sum ac- tually expended ; in Brokerage, upon the par value, or an as- sumed value. III. Ex. My agent buys a quantity of goods for $220 ; what is bis commission at 5 % ? 1220 X .05 — 111, Ans. Or, 5 % :i= 2^0 ; j(T of ^220 = $11, Ans, Examples. 1. What should a commission merchant receive for selling 4750 pounds of sugar at 12^ cents a pound, his commission being 1% ? " Ans. 5.937+. 2. A stock broker purchases for a person 8 shares of stock in a manufacturing company at $72 a share ; what is his commission ati%? 3. What is a broker's commission for negotiating a loan of $4500 at ^%? 4. Dupee & Sayles bought on account of T. Winship, 4 sharee 210 PERCENTAGE. of EjiSex Company's stock, at $27 each, their commission being ^% ; what is Winship's bill ? Ans. $108.27. 5. What amount of current money will be given in exchange for $450 of that which is at 5% discount ? Ans. $427.50. 6. When gold is at a premium of 25%, what must be paid for $275 of gold? Ans. $343.75. 7. If an auctioneer sells, on a commission of 8%, 14 chairs at $1.25, 1 bedstead for $10, and a miscellaneous lot for $53.79, what sum will be due the person for whom he makes the sale, his commission being deducted ? 8. What is the commission on the sale of 200 yards of broad- cloth at $4.80 per yard, 6% being paid for selling, and 2^% for guaranteeing the sales ? Ans. $81.60. Note. — The seller sometimes guarantees the payment for the goods sold ; for this he is paid a premium. 9. AVhat are the net proceeds on the sale of a lot of crockery amounting to $10650, commission being 4^%, and lJ-% being allowed for guaranteeing payment? Ans. $10011. Note. — To obtain net proceeds, deduct commission. 10. What are the net proceeds from the sale of 1260 barrels flour, at $3.50 per bbl., charges for freight and storage being 40 c. per bbL, commission for selling being 2%, and for guaranteeing sales 1^%? 11. What are 50 shares of railroad stock worth, at 4% ad- vance, the par value being $100? $100 X 1.04 X 50 =r $5200. Ans. 85200. 12. What would be the value of 15 shares of the above stock, at 7 ^c premium ? 13. What would be the cost of 8 shares of the above, at a dis- count of 35 % ? 14. What would be the value of 4 shares in the stock of a gas company, originally Mwth $200, at 3 % above par ? 15. What would be the value of 12 shares of above stock, at 17% below par? 16. A certain corporation, wishing to increase their capital COMMISSION AND BROKERAGE. 211 Stock Without multiplying their shares, assessed the stockholders 40% on the par value of their stock, which was $500 per share ; what was assessed on a person holding 3 shares ? 17. What was the par value of the stock per share in the corporation after the assessment was made ? 18. If I buy 10 shares of stock, originally worth $100, at 18% above par, and sell it at 7% below par, what do I lose ? Ans. $250. 19. What would have been my gain if I had bought the above at 10% discount, and sold it at a premium of 8% .-* 20. Bought 75 shares in a savings-bank, par value being $50, at 61% advance, and sold at 3^% above par; what did I lose on the lot ? 21. The amount of the deposits in the savings-banks of Massa- chusetts for 1863, was $44,785,438.56; the ordinary dividends w«re at the rate of 4|% of the deposits; what was the total of the dividends ? 301, To FIND THE Commission or Brokerage and the Sum invested, when both are contained in a cer- ^iN Sum sent to a Factor or Broker. III. Ex. I send to my agent at St. Salvador $1224; what part of this sum will remain to invest in sugars, after deducting his commission of 2 % on what he lays out ? Operation. Since the commission is $1224-^-$1.02 — $1200, sum to invest. 2^ of the sum laid out, $1224— $1200 zzz $24, commission. the agent receives $1.02 for every dollar which he is to lay out. If he receives $1224, he will have as many dollars to lav out as $1.02 is contained times in $1224, which is 1200 times. ^ Ans. $1200. Hence the . Rule. To find the sum invested : Divide the amount named hy $1 plus the commission on $1 ; the quotient will he the sum invested. To find the commission or brokerage : Subtract the sum in^ vested from the amount. 212 PERCENTAGE. Examples. 1 I have sent to a commission merchant in New York $450, of which he is to lay out what he can in butter, after reserving his commission of 2 % on the purchase ; what is the purchase ? Ans. ^441.176+. 2. What part of a remittance of $328.25 will remain to be in- vested after 1 % of the investment has been deducted ? Ans. $325. 3. How many barrels of flour at $5 each can a factor pur- chase with a remittance of $2575, after deducting his commission of3%? Ans. 500 hh\. 4. How many shares of stock at 1 100 each can a broker pur- chase with a remittance of $520, allowing himself a brokerage of 4%? 5. A real estate broker receives $2593.75 ; what number of acres of land at $1.25 per acre can he purchase with the balance after allowing himself 3| % brokerage on the purchase ? 6. Having remitted to my agent in New Orleans $891.75, to be expended for sugars, after reserving his commission of 2^-%^ I received from him 29000 pounds of sugar ; what was the cost per pound ? 7. I have authorized a broker to employ $292.32 in the pur- chase of a certain stock for me, allowing him 1^% commission; what number of shares originally worth $100 can he purchase, if they are now 72 % below par ? 8* Wm. H. Ladd sells for Chas. Smith 2500 pounds of butter at 14 cts., and 100 pelts at $1.50 ; from the proceeds he deducts his commission of 3% and $4 for cartage, &c., and with the bal- ance purchases for Smith, after deducting his commission of li% on the purchase, a lot of sheeting at 10 cts. per yard; how many yards can he purchase ? i^^ Tor Dictation Exercises, see Key. INSURANCE. 213 INSURANCE. 309. Insurance is security to indemnify for loss. Property Insurance indemnifies for loss by fire, shipwreck, &:c. Life and Health Insurance indemnify for loss of life or health. 303, The persons or company that insure are called'under- writers. 304:. A Policy is the written contract between the insurer and the insured. 30^. Premium, is a certain per cent, of the sum insured paid to the underwriters for the insurance. 300, Examples. 1. Required the premium for insuring a house for $1600 at ^%. 1% of $1600 = $16; ^% of $1600 = $8, A?is. 2. What is the insurance on $1000 worth of furniture at ^%, including $1 for policy ? Ans. $6, 3. Insured | of a store valued at $15000 at |% jand paid $1 for policy. What amount is paid ? 4. Effected insurance on the ship Susan to Cadiz and back for f 10000 at 2%, and on her return cargo, worth $7500, at 1^% ; what is the amount of insurance, including $1 for policy? Ans. $313.'b0. 5. A lot of clothing worth $4000 is insured for § of its value at I % ; if tlie goods are damaged by fire to the amount of $500, what will be the net loss to the underwriters, making no account of interest ? Ans. $480. Note. — The underwriters will make good to the insured his actual loss. Their net loss will be $500 minus the premium. 6. What will be the premium for insuring $15500 on a school- house for 10 years at 2|% ? 7. What would be the loss to the insurance company if the 214 PERCENTAGE. above building should be destroyed by fire, and the insurance be paid in 6 months from the date of the policy ? Ans. ^15116.84. Note. — Reckon interest on the premium for 6 months. 8. Insurance was effected upon |- of a ship and cargo, valued at 85000G, at lg% ; what would be the actual loss to the under- writers if the ship and cargo should be totally lost at sea, making no allowance for interest ? 9. What would be the actual loss to the owners ? A?is. $13203.120. 307, To FIND FOR WHAT SuM AN INSURANCE POLICY MUST BE TAKEN OUT, TO SECURE A CERTAIN SuM AND COVER THE Premium. III. Ex. For what must a policy be taken out to insure $500 on a ship's freight, and cover the premium of 2% .'^ Operation. 1 — .02 = .98 Since the premium is 2-% of the policy, the $500 -J- .98 ==$510,204. property ($500) must be 98^ of the policy; if $500 is 98^, 1^ will be Jg of $500, and 100^ will be 100 X gV of $500, which is $510,204. Ans. $510,204. Hence the Rule. To find for what sum an insurance policy must be taken out, to secure a certain sum and cover the premium : Di- vide the sum to be secured hy 1 minus the rate per cent, of insur- ance ; the quotient will he the amount of the policy. • Examples. 1. "What policy will cover $2000 of property and a premium of3%? Ans $2061.855+.' 2. I have loaned $1140 to a friend, to be secured by a policj* on his life ; for what must a policy be taken to secure the sun? loaned and cover the premium of 5% also ? Ans. $1200 3. Having adventured $1800 to Calcutta, what policy shouh' I take>put to secure both the adventure and the premium of 6%-' 4. For what must a policy be taken out to cover a loan of $588 and a premium of 121% upon it? i^" For Dictation Exercises, see Key. EQUATION OF TAYMENTS. 215 AVERAGE, OR EQUATION OF PAYMENTS. 308. Equation of Payments is the process of finding an average time for the equitable payment of several sums due at different times. 309. The Equated Time is the date at which all the items may be paid without loss to either party. 310. The Term of Credit is the time from the contract- ing of a debt to the date of its becoming due. 311. To FIND THE Equated Time, when the Terms op Credit begin at the same Date. III. Ex. I owe P. Benjamin two notes dated March 1, — . one for $80, to be paid in 12 months, the other for $40, to be paid in 3 months. "When, without loss to either Benjamin or myself, can I pay both notes at once ? Interest Method. Operatiox. I am entitled to keep these 12 months' interest on $80 =: $4.80 two notes till their interest 3 " " " 40 i=z .60 equals $5.40 ; if I pay them both Y^Q g ^Q at once, it should be at such time 1 of .1^ of 120 = .60 ; .60 ) 5!40 ^^^^'' ^^^''' ^ ^^ '"^'^^ ^^ ^^^^^^^^ ^ ^°^ for $120 to gain $5.40; $120 gains 60 cts. a month; .*. to gain Mar. 1+9 mo. z=z Dec. 1, Ajis. *- ^a v mi ' ^0.40, it will require as many months as 60 cts. is contained times in $5.40 = 9 months, which added to Mar. 1, is Dec. 1. Hence KuLE I. To find the equated time when all the terms of credit begin at the same date: Find the interest on each item for its time of credit ; divide the sum of the interests hy the in^ terest of the sum of the items for one month. The quotient wilt he the equated time in months. Add the equated time to the date. Note I. — To obtain the interest for 1 month, remove the desimal point two places to the left, and divide by 2. 216 PERCENTAGE. Note II. — If any item contains cents, reject tliem if they are less than 50, and increase the dollars by one if they equal or exceed 50. Product Method. Operation. The use of $80 for 12 m. = the 80 X 12 = 960 ^^gg of |1 fQ^. ggQ ^^ . ^j^g ^gg ^^ ^Q X ^ =^120^ $40 for 3 m. = the use of $1 for 120 ) 1080 ^^^ ^' ^^ ^°^' ^^^ ^- + ^^ ^^^ 120 m.=:$l for 1080 m., M^hich^ 9 m. $120 for ^\-^oi 1080 m. or 9 m., which added to Mar. 1. zzrDec. 1, Ans. Hence Rule II. Multiply each payment hy the number of days or months to elapse before it becomes due ; divide the sum of the prod- ucts hy the sum of the payments^ and add the quotient to the date. Note. — The examples in this book are performed by the Interest method. Examples. —1. What is the equated time for paying $50 due in 5 m. from May 14, 1863, |35 due in 4 m., and $25 due in 2 m. from the same date? Ans. Sept. 14, 1863. ^% B. Frank holds five notes against me, dated June 7, 1864 ; one for |500 on 4 m., one for $750 on 5 m., one for $200 on 12 m., one for $400 on 2 m., and one for $400 on 17 months* credit ; what is the time at which all should be paid if paid in one sum? Ans. Jan. 7, 1865. -^. Having sold Samuel Bond real estate to the amount of $2000, lie gave me four equal notes for it, dated Oct. 4, and pay- able in 5, 6, 9, and 12 months ; what is the average time for the payment of all the notes ? Ans. 8 m. i^A. What is the average time for paying $20 due in 20 days, $20 due in 100 days, $70 due in 30 days, $20 due in 60 days, and $40 in 70 days ? Ans. 1 m. 20 d. -6. April 1, C. A. Brown purchased coal to the amount of ^5000, -^ of which was to be paid in 6 months, ^ in 12 months, EQUATION OF TAYMENTS. 217 and the remainder in 9 months ; for what time should a note without interest, dated April 1, 1865, in payment of all the sums, be allowed to run, and when should the note be paid ? Ans, Jan. 28, 1866. ~^a Aowes B ^360.25 (Note 2, ArtSll),^ of which is to be paid in 7 months, ^ of the remainder in 8 months, ^ of what then re- mains in 10 months, and the balance in 4i months; in how many- months and days should the whole be paid J' • Ans. 6 m. 22 d. 7. Sept. 25, bought 3 parcels of goods, as follows : a bill amounting to $225.25 on 12 months' credit, a bill amounting to $125^.25 on 8 months' credit, and a bill amounting to $40 on 5 months' credit; what was the mean time for paying all? Ams, 10 m. Note, — "When a sum. is paid immediately, the term of credit expires instantly, and it will have no corresponding interest or product in time. 8. A person promised to pay $7000, |- of it immediately, ^ of the remainder in 8 months , ^ of what then remained in 22 months, and the balance in 16 months ; what is the equated time for paying the whole ? Ans. 12 m. ~~~9. A merchant tailor finds, on examining his account with Jones <& Ca, May -5, that he owes them for 150 yds. of silk, at $.50 per yd., which is due that day ; for 2339 yds. of cambric, at $.10, which will be due in 6 days; for 12 J- yds, broadcloth, at $5.00 per yd., w^hich will be due in 1 6 days ; for 50 yds. of doeskin at $3,75 per yd., which will be due in 20 days ; wliat is the average time for paying the whole ? If the tailor settles the account by giving his note, when should the note begin to bear interest? 313.). To FIND THE E<5UATED TiME, WHEN THE TeRMS OF Credit begin with different Dates. III. Ex, J, Rives bought of A. Ainger the following bills of goods : — Sept, 15, a bill amounting to $100, « 30, « « *< $400, 218 PERCENTAGE. Oct. 8, a bill amounting to $250, " 10, " " " $250, ■ What is the equated time for paying the whole ? To equate the above bills, it is necessary to assume a date from which to compute the interest on the several items ; any date may be assumed, but the most convenient date for examples generally, on account of reckoning the time, will be found to be the last day of the month before the earliest date at which any item becomes due ; this in the above example is Aug. 31. OPERATION BY INTEREST METHOD. g j ^^ ^j^e assumed ^ate, 15 days'mterest on$100:=: .25 » oi -d- ij i ♦. [ Aug. 31, Hives would lose mterest on the several bills from Aug. 31 400=2.00 ^Q „ ,, ,, 250= 1662 to their respective dates, amount- ■ _l_Jl ing in all to $5.50; .*. payment ^^^^ ^-^^ should be made at such time after ^ofyl^of 1000z=5; 5) 5.50 Aug. 31 as will be required for J^ $1000 to gain $5.50, = 1 m. 3 d., 1.1 m. =: 1 m. 3 d. which added to Aug. 31 is Oct. 3, Aug. 31+1 m. 3 d. rr Oct. 3. -^ns. Hence the Rule. To find the equated time when the terms of credit begin with different dates : Assume that the time for paying all the items is on the last day of the month previous to the earliest day at which any item is due ; find the interest on each item from the assumed date to the date when it is due, and divide the sum of the interests by ^ of-j-hu (fihe sum of the items ; the quotie^it will be the time after the assumed date^ in months, when all should be paid. Note I. — .1 month = 3 d. ; .03 J month = 1 day nearly, etc. Note II. — Reject the fraction of a day when it is less than -| ; reckon it 1 day when it is ^ or more. "^1. Find the equitable time for the payment of the following' $300, due April 1, 1869; $450, due Dec. 1, 1869; and $600, dutf March 10, 1870. (Assumed date, March 31, 1869.) Ans, Nov. 25, 186^ EQUATION OF PAYMENTS. 219 ^. Find the mean or average time for paying the follow- ing: S12.45, due Feb. 10, 1860; $24.17, due Mar. 1, 1860; $15, due Mar. 14, 1860; $30, due Mar. 16, 1860; and $12.70, due Mar. 7, 1860. Ans. Mar. 5, 1860. Edwin Foote's ledger contains the following account. Thomas Wing, Dr. Or. 1861. July 1. To Merchandise. 250 00 1862. Apr. 1. (( i< 400 00 Oct. 2. « {( 600 00 Note. — This account shows that Wing bought of Foote merchandise at the times and to the amount indicated. ,.t 3. Allowing interest on each item from its date, what is the time from which a note should draw interest in payment of all of the above items ? Ans. May 9, 1862. 4. What is the equated time for paying the following bill ? New York, Jan. 1, 1861. E. Train & Co. 1860. Jan. 20. Mar. 15. Apr. 12. Bought of F. Fogg & Co. M'd'se on 3 mo., $100. « " 2 " 100. " " 2 " 100. The above items will be due as follows: April 20, May 15, and June 12. Equate from these dates. Assumed date. Mar. 31. Ans. May 16, 1860. 5. Equate the following : — RoxBURY, Jan. 1, 1865. Mr, J. Stow 1864. Bo't of Z. Churchill, Jan. 5. M'd'se on 5 mo., $400. May 5. « " 4 " 600. u 16. « a 4^ ^ 200. Ans. Aug. 8. 220 PERCENTAGE. 6. If one note should be given for the following three, when should interest commence upon it ? A note for $200, dated May 15, 1864, on 90 days. " " " 250, " June 1, " " 60 « « « " 700, « July 8, " « 30 " rh^i*^'^ 7. "What is the mean time for the payment of the following bills of goods purchased by Calrow & Co. of Armstrong & Co. ? 1856. June 1, a bill of $200 on 90 days. « Feb 1, « « 800 " 75 " « Apr. 1, « " 300 « 60 " " July 1, "■ " 650 « 40 « ^P 1857. Feb. 1, " " 1000 « 20 «, (W^t /- 8!" What is the equated time for paying the following ? Boston, Juli/ 1, 1864. J. P. Putnam, To Weymouth Iron Co., Dr. -\H(c^ liJ 1S64. Jan. 10. " 28. Feb. 29. Mar. 12. ApB. 8. To Merchandise on 3 mo., " " " 3 mo., " " " 60 d., " " " 4 mo., « « " 90 d., $437 254 144 159 I 300100 9t Find the equated time for the payment of the following notes held by Page & Son against AYashington Manufacturing Co. A note for $560. dated Jan. 1, 1856, on 5 r Qon a a 846.15 a Feb. 11, " '* 6 a a a 728.50 li Mar. 20, " « 6 a u (( 400. a July 30, « " 6 (( *( u 560. a Sept. 12, " « 8 a n u 600. a Dec. 18, « « 6 i( « n 500. a May 10, 1857, (I a a (( « 350.75 i( « 7, « i( a a a (( 820.20 a Apr. 17, " (( a a ^T For Dictation Exercises, see Key. AVERAGE OF ACCOUNTS. 221 AVERAGE OF ACCOUNTS. 313. To Average an Account. III. Ex. I have in my ledger an account with F. E. Clarke, both the debt and credit sides of which consist of sundry items. The footing is as follows : — Dr. side $250, averaging due Feb. 9, 1863 ; the Cr. side $300, averaging due Apr. 4, 1863 ; at what time should I pay Clarke the balance ? Assuming, as in Art. 312, Jan. 31, 1863, as the time for settling this account, we compute interest on each item from this time till it is due. By settling Jan. 31, 1863, I should lose 63 days' int. on $300 = $3.15 Clarke would lose 9 days' int. on $250 z= $.375 The balance due Clarke is $50. $2,775 int., my net loss. i of ^^ of $50 = $.25. 2.775 -^ .25 = 11.1 m. = 11 m. 3 d. Jan. 31, 1863 + H m. 3 d. ;= Jan. 3, 1864. If, by settling at the assumed date, my net loss of interest would be $2.775, 1 shall be entitled to keep the balance, $50, till it has gained $2,775, which, found by dividing it by the interest of $50 for 1 month, is 11 m. 3 d.; this added to Jan. 31, 1863, is Jan. 3, 1864, Ans. If, however, the dates were transposed, making Clarke's $250 due to me Apr. 4, and my $300 due to Clarke Feb. 9, by settling at the assumed date Clarke would lose 63 days' interest on $250 ■= $2.62^, I should lose 9 days' interest on 300 z= .45 The balance due Clarke is $50. $2.17^ int., C.'s net loss. If, by settling at the assumed date, Clarke's net loss of interest is $2.17^, he may justly require me to pay the balance, $50, at such time prior to Jan. 31, 1860, as will be required for $50 to gain $2.17^ of interest, which is 8 m. 21 d. ; this, reckoned back from Jan. 31, 1863, is May 10, 1862. From the above example we deduce the following 222 PERCENTAGE. Rule. To equate an account : Assume that all the items art to he paid on, the last day of the month previous to the earliest day at which any item hecomes due; Jind the interest on each item from the assumed date to the date at which it becomes due ; find the difference between the interest on the Dr. and Cr. sides of the account; divide this difference by the interest on the balance of the account for 1 month ; add the quotient to the assumed date when the larger side has the more interest, and subtract it from the assumed date when the larger side has the less interest. 314. Settlement can be effected earlier than the equated time, by deducting interest from the balance of the account for the time between the equated time and the desired time of set- tlement. It can be effected later by adding interest. The latter will be necessary when the equated time is already past. Or, 31^. If it be desired to settle an account at a specified time, add interest to the items due before the specified time, and subtract interest from those due after the specified time ; the dif- ference between the total of the Dr. and Cr. items plus or minus their interest, will be the balance due. 310* Examples. 1. When can the balance of the following ledger account be paid without loss to either party ? Dr. Edward C. Damon. Cr. 1863. Apr. 1 July 8 To Cash, $1000 ^' Mdse., 118 00 98 1863. Apr. 14 Aug. 10 By Mdse., " Real Estate, $1392 59 94 33 This mon re( Apr. July account shows that leives, 1, $1000. 8, 118.98. Da- And April Aug. that he is credi 14, with $139 10, « 9 ted, 2.59. 4.33. We assume March 31 for settling the account. AVERAGE OF ACCOUNTS. 223 Data. Apr. 1, July 8, Operation. Br. Cr. mem. Days. Int. Date. Item. Days. $1000 1 $ .166f Apr. 14, $1393 14 119 99 1.9635 Aug. 10, 94 132 Int. $3.25 2.068 1119 2.130+ 1487 1119 5.318 2.130 Balance of %, $368 Bal. of int. ^^ of $368 = 1.84 ; 3.188 +- 1.84 = 1.73 = 1 m. 22 d Mar. 31 + 1 m. 22 d. =z Ans. May 22, 1863. $3,188 2. Add 3 months' credit to each item in the following, and equate the %. Dr. a. B. in % with C. D. Ck. 1863. Jan. 1 Mar.31 May 30 To Bal Ledger B., " Heal Estate, " do. do. $50 50 55 00 00 00 1863. Mar. 3 May 27 By Mdse., $50 50 00 00 Assumed date, Mar. 31, 1863. Ans. May 12, 1863. ^. When is the balance of the following ^ due .'' Dr. Smith, Dove, & Co- Cr. 1862. Jan. 6 Feb. 7 1862. [dse.,30d.er., $600 00 Jan. 1 do. 60d. cr., 840 Mar. 16 By R. Estate, 90d.cr., " Cash, $500 300 Assumed date Jan. 31, 1862. Atis, Feb. 25, 1862. 4r When is the balance of the following account due ? Db- Day, Wilcox, & Co. Cr. 1864. July 21 To Mdse. 90 da.. Aug. 15 " do. 60 da., Aug. 31 '' do. 4 mo.. Oct. 17 '' Cash, $173 13 81 230 00 1864. June 25 June 30 Aug. 20 Sept. 12 By Mdse. , 30 da.. $500 00 " do. 60 da., 52 71 " do. 4 mo., 16 48 " do. 30 da., 102 10 A71S. Dec. 26, 1863. 317, To find the equitable time for the payment of the bal- ance of a debt, when partial payments are made before the del/' 224 PERCENTAGE. is due : Make the whde debt the Dr. side of an account, and the partial payments the Or. side. 5. A holds a note against B, dated Nov. 14, 1864, for $620, due 7 months hence, without interest. On this note B paid A $220 Jan. 14, 1865, and $100 Feb. 14, 1865 ; what is the equated time for paying the balance? Ans. Nov. 16, 1865. 6. T. Ropes owes R. Treat $250, due May 29. If he should pay $50 on the 29th of April previous, when should he pay the balance ? Aiis. June 7. 7. A fanner purchased, on the 1st day of April, 1864, 3 acres of land at $183 per acre, agreeing to pay for it in 7 months; if he should pay $50.75 at the date of the purchase, $148.25 in 4 months, and $150 in 3 months, in what time should he pay the- balance ? i- j u 8. A owes B $2000 Oct. 5 ; if he should pay $1200 of it Sept. 8, at what time should the balance be paid ? 9f J. Edwards owes J. Adams $1200 on a note dated Oct. 9, 1863, payable in 4 months without interest ; if Edwards should pay Adams $250 on this note Jan. 16, 1864, and |400 Feb. 9, 1864, when should the balance of the note be paid? 10. Bought a lot of land for $800, for which I gave my note, dated May 7, 1864, payable in 6 months ; June 28, I paid $158 ; Aug. 7, 1 paid $320.60, and Sept. 7, $179.40; when should the balance be paid ? lit Find the equated time for the settlement of the following account : — Robertson & Reynolds in % with James Loring & Co. Dr. Cit. ~1864. 18^. July 12. To Balance, $562 IT July 18. By Cash, $480 00 " 20. " Mdse. on 4 mo. 1524 82 " 27. " Note on 90 d. 1218 65 Aug. 8. " " " 2 mo. 210 00 Aug. 20. " Real Estate, 600 00 Sep. 30. " " " 4 mo. 783 25 Sep. 30. " Cash, 459 50 Nov. 25. " Bill due, 286 58 Oct. 28. "Draft at eOd.f 425 00 Dec. 1. (( it (( 424 60 1 iDec. 1. " Cash, 185 20 ^p* For Dictation Exercises^ see Key. t Allow S days of grace. TAXES. 225 TAXES. A Tax 13 a sum of money assessed upon a person or upon property for public purposes. 319. A Poll Tax is a sum assessed upon each male citizen liable to be taxed, without regard to his property. The persons thus taxed are called the polls. 3^0« Real Estate consists of immovable property, such as houses, lands, &;c. SSI* Personal Property consists of movable property, such as money, stocks, cattle, ships, &c. 3SS. Assessors are officers appointed to levy taxes. It is their duty to ascertain the value of the taxable property and the number of polls, and to apportion the tax to each person. 3S3. III. Ex. The whole state, county, and town tax of Oxford for the year 1862, was $5300 ; the value of the real es- tate and personal property is $1250000; there are 200 polls in the town, each of which is taxed $1.50. What is the tax on $1, and what is J. Swan's tax, who has $3000 of real estate anc^ $1000 of personal property, and who pays 1 poll tax ? Opkration. $1.50 X 200 = $300, amount of poll taxes. $5300 — $300 := $5000, property tax. $5000 -^- $1250000 =^ 4 mills, tax on $1 of property. $3000 -f $1000 ZI3 $4000, Swan's taxable property. $4000 X $-004 — $16.00, Swan's property tax. $16.00 + $1.50 z=z $17.50, Swan's whole tax. Hence the Rule for Apportioning Taxes, Multiply the tax on one poll hy the number of polls, and subtract the product from the whole tax ; divide the balance by the taxable property i the quO' tient is the tax on $1. Multiply each person's taxable property by the tax on $1, and add his poll (ax^ or taxes^ if he have any. 226 PERCENTAGE. Examples. 1. The whole tax of the town of H. is $70352 ; the valu- ation of the town being $9329000, the number of polls being 3366, each taxed $1.50, what is the tax upon $1 and what is the tax on the following named tax-payers ? A has property amounting to $8500, and pays 1 poll. B has property amounting to $3570, and pays 1 poll. C has property amounting to $5800, and pays polls. D has property amounting to $1000, and pays 2 polls. E has property amounting to $2800, and pays 3 polls. Ans. 7 mills on $1 ; A's tax, $61 ; B's tax, $26.49. 2. The town of L. votes to raise a tax of $8343.20 ; the val- uation of the town is $2000000 ; there are 1679 polls, each taxed $.80 ; what is the tax on a dollar, and what is the tax of J. L. Partridge, who has $1500 of real estate and $3000 of personal property, and pays two poll taxes ? A71S. 3i mills on $1 ; Partridge's tax, $17.35. 3. What would be the tax upon a non-resident who had prop- erty in the above-named town of L. to the amount of $15225 .-^ Ans. $53,287+. 4. The state tax of a certain town is $3093 ; the county tax is $5110; the town tax, $33860; the valuation of the town is $6700000 ; there are 2542 polls, each taxed $1.50. What is the tax on $1, and what is the tax on a person having $4500 in real estate, $2750 in personal property, and who pays one poll tax? Ans. $.005^^4 on $1 ; $42.889ff. >6. There is a town whose valuation is $1100000, in which there are 300 polls, each taxed $1.20 ; the tax to be raised is $9600. What is the tax on $1, and what is the tax of a person having $4000 in real estate, an annual income of $3000, on all above $800 of which he is taxed as for personal property, and who pays three poll taxes ? Ans. 8| m. on $1 ; $55.68. 6. The valuation of a certain town in real estate is $3200000 ; in personal property $1186000 ; the tax to be raised is $31579.20, TAXES. 227 one sixth of which is to be levied upon the polls, of which there are 3096. AVhat is the tax on $1, and what on each poll? Ans, G m. on $1 ; poll tax, $1.70. Assessors commonly construct a table showing the tax on $1, 72, $3, &c.; from which they compute the individual taxes. 7. The valuation of a certain town is $11522400; the tax to be raised is $108391.GO ; there are 3350 polls, each taxed $1.40. Find the tax on $1, and perform the remaining examples by the following Table, Showing the tax on various sums at the rate of $.009 on $1. $1 pays $.009 $10 pays $.09 $ 100 pays $.90 2 " .018 20 u .18 200 u 1.80 3 " .027 30 a .27 300 « 2.70 4 « .036 40 « .36 400 u 3.60 5 « .045 50 « .45 500 « 4.50 6 " .054 60 it .54 600 (( 5.40 7 « .063 70 (( .63 700 (t 6.30 8 « .072 80 (( .72 800 u 7.20 9 « .081 90 a .81 900 (t 8.10 10 " .09 100 « .90 1000 t( 9.00 8. At the above rate, what is A's tax, he being assessed for $4250, and paying 2 polls ? Operation. $4000 pays $36. 200 '♦ 1.80 50 " .45 2 polls " 2.80 Ans. $41.05, A's tax. Find the tax of the following at the above rate : — t 9. Of B, who is assessed for $2800 and 1 poll. 10. Of C, who is assessed for $7850 and 3 polls. 11. Of D, who is assessed for $1565 and 1 poll. 12. Of E, who is assessed for $906^ and 2 polls. 228 PERCENTAGE. 13. Of F, who is assessed for $6555 and 1 poll. 14. Of G, who is assessed for $5687 and 1 poll. 15. Of H, who is assessed for $10793 and 3 polls. 16. Of I, who is assessed for $3384 and 1 poll. 17. Of J, who is assessed for $4597 and 1 poll. ' 18. Of K, who is assessed for $8979 and 2 polls. 19. Of L, who is assessed for $2972 and 1 poll. 20. Of M, who is assessed for $1000 and 1 poll 21. Of N, who is assessed for $6587 and 2 polls. 22. Of O, who is assessed for $7572 and 2 polls. 23. Of P, who is assessed for $2956 and 1 poll. ^^ For Dictation Exercises, see Key. CUSTOM HOUSE BUSINESS. 334:* Custom Houses are places where Government Offi- cers collect duties. 3S^. Duties are taxes upon imports and upon the tonnage or weight which a vessel may carry. They are of two kinds, specific and ad valorem. They furnish a revenue for the gov- ernment. 3S6. An Invoice is a list of imported goods, showing their quantity and price. 3^7 • A speeiflo duty is a tax upon each article of a cer- tain kind, without regard to its value. 328. An ad valorem duty is a certain per cent, of the cost' of goods, estimated upon the invoice. 3^9* Leakage and Breakage are allowances for loss from the leaking and breaking of bottles, boxes, (Sec. 330* Tare is an allowance for the weight of boxes, &;c, 331. Gross Weight is the weight of goods including what- ever is used for packing. 33^, Net Weight is the weight of the goo<^s alone. CUSTOM HOUSE BUSINESS. 229 333. Examples. 1. What is the net weight of 120 boxes of raisins, gross weight being 31 lbs. each, the tare being 6^ lbs. per box ? Ans. 2940 lbs. Operation. 31 — 6^ == 24|, net weight of 1 box. 24^ X 120 — 2940, « " " 120 boxes. 2. What is the duty, at 5 cents per lb., on the net weight of the above ? Ans. $147. 3. What is the specific duty, at 15 cents per gallon, on 25 bar- rels spirits turpentine, containing 32 gallons each, 5^ being deducted for leakage ? Ans. $114. 4. What is the duty, at 25 %, on 75 boxes of tin, 112 lbs. per box, invoiced at 7 cents per pound, tare being 6 lbs. per box? Ans. $139,125. 5. What is the duty on 100 dozen watch crystals, at 35 5^, in- voiced at $1 per dozen, 3% being allowed for breakage? Ans. $33.95. ^ 6. What is the duty, at 20 % ad valorem, on an invoice of 24 boxes of tea, gross weight being 1284 lbs., 8 lbs. for tare being allowed on each box, the cost of the tea being 38 cents per pound? Ans. $82,992. 7. I have imported 3 tons, 3 cwt. 3 qrs. 7 lbs. of steel invoiced at 20 cents per pound ; 8 % being allowed for damage, what is the. duty at 20%? t 8r What is the cost at the store of 5 hhds. of sugar, weighing gross 255 G lbs., which was bought in Havana for $178.92, and on which is paid $35.75 for freight and carting, and 2^ cents per pound for duty after deducting 15 % for tare ? h~ — ^ d* Find the cost of two cases of gum arable, at 21£. 5s. per cwt., duty 30 % ad valorem ; the gross weight of the first being 1 cwt. 3 qrs. 20 lbs., of the second 1 cwt. 1 qr. 10 lbs., 35 lbs." being allowed for the weight of each case. Ans. 7d£ Os. 2fd. I^^ For Dictation Exercises, see Key. * See Art. 1G4, Note. 230 PERCENTAGE- DRAFTS AND BILLS OF EXCHANGE. 334. A Draft is a written order, directing one person U pay money to another. The following is a simple form of a Draft. $100. Baltimore, April 4, 18G4. Thirty days after sight, pay to Samuel Price, or order, One Hundred Dollars, and charge the same to my account. Charles Smith. To. Brewer & Tileston, Publishers, Boston. 33^. The Drawer is the person who signs the draft. 336. The Drawee is the person to whom the draft is -iid' dressed. 33 T. The Payee is the person in whose favor the draft 5:* drawn. Note. — In the above, Charles Smith is the Drawer, Brewer & Tileston are the Drawees, and Samuel Price is the Payee. 338. The Holder of a draft is the person who has legal possession of it. 339. The Endorsement of a draft is the writing upon its back, by which the payee transfers his right in it to another person. 34:0. If the drawee does not pay the money when the draft is presented, he may signify his acceptance of it by writing his name on its face, after the word " Accepted," by which act he becomes responsible for its payment. Bills of Exchange. 341. A Bill of Exchange is a draft used by merchants for the discharge of debts payable at a distance. DRAFTS AND BILLS OF EXCHANGE. 231 34^* Bills of Exchange are Inland, or Domestaic, when they are drawn and payable in the same country. 34LS. Bills of Exchange are Foreign when they are drawn in one country and payable in another. Illustration. — Suppose A, in New York, ships butter to B, in Liv- erpool, to the amount of $1000. He makes a draft on B to pay to him- self, or " bearer," or " order," the $1000 due. But C, in Boston, wishes cutlery from D, in Sheffield, to the same amount. So he buys A's draft, paying its value in United States money, and sends it to D. D receives it and presents it to B, in Liverpool. B, having received his butter in good condition, accepts the draft, and when the time comes for pay- ment of the money, pays it to D, of Sheffield, in English currency. Thus A receives payment for his butter, and D for his cutlery, without the risk or inconvenience of sending the money from one country to another. 34:4:. Bills of Exchange thus bought and sold are ^aid tu be negotiable, or marketable. •S4:«i» When the value of the goods sent from the United States to another country — England for example — is greater than the value of those received from England, more money is due to us from the English merchants than is due to them from our merchants, and we hold more bills against England than are needed to pay our debts ; consequently, bills become cheap, and exchange is at a discount. When the value of the goods im- ported from England exceeds the value of those sent to England, our merchants hold fewer bills against England than are needed to pay their debts, and bills thus become dear, and exchange is at a premium. The current price of exchange is called the Course of Exchange. 34:6, A Set of Exchange consists of two or more drafls of the same tenor and date, the payment of either one of which cancels the other one or two. To provide against accident in the transmission of a drafl, i\ ia customary to send two, at least, of a set by different model ©f conveyance, or at different times. 232 PERCENTAGE. INLAND OR DOMESTIC EXCHANGE. 34L7* To FIND THE Value of an Inland Draft. III. Ex., I. "What must be paid in St. Louis for a draft on Philadelphia, for $2500, payable at sight, exchange being 1^% premium in favor of Philadelphia ? Operation. If exchange is 11^ premium, $2500 X 1.015 = $2537.50, Ans. the exchange value of $1 is $1,015 and the price of $2500 will be $2500 X 1.015 r= $2537.50, Ans. III. Ex., II. What must be paid in New Orleans for a draft on New York for $1500, payable in 60 days after sight, exchange being 1% premium ? Operation. $1500 ^1.01 =: $1515., Exchange value of $1500 at sight. $1500 X .0105 = 15.75 , bank discount of $1500 for 63 days. Alls. $1499.25, cost of draft. Hence the Rule. To find the value of an inland draft : Multiply the face of the draft hy the exchange value of %1. If the draft is payable after sight., deduct from the product the hank discount of the face of the draft for the given time and grace. Examples. 1. What is the value of a draft on Boston for $2500, when ex- Change is at a premium of | % ? Ans. $2506.25. 2. What must I pay for a draft on New York for $700 at 12 days without grace, exchange being at |% premium? Ans. $699,475. 3. Bought a bill on New Orleans for $400, at a discount of 1 % ; what did I pay ? Ans. $398. *^4. Messrs. B. & T., of Boston, sold a draft for $206.59 on Billings & Son, of Baltimore, at 30 days' sight, discount being ^% ; what did they receive for it? EXCHANGE. 233 S4:8# To FIND THE Face op a Draft. III. Ex. What is the face of a 30 days' draft on Cincinnati at 1 % discount, which can be purchased at New York for $200 ? Operatiox. $1 — $.01 =1 $.99, exchange value of $1 at sight. .0055, bank discount of $1 for 33 days. .9845, exchange value of $1 at 30 days. $200 -^ .9845 =: $203,148+, Ans. The exchange value of $1 of the draft will be $.9845 ; if $1 can ho purchased for $.9845, as many dollars can be purchased for $200 as $.9845 is contained times in $200, which is 203.148-}- times. Ans. $203,148-1-. Hence the Rule. To find the face of a draft which may be purchased for a given sum : Divide the given sum ly the exchange value of n. Examples. 1. Wishing to remit to my correspondent at St. Louis the net proceeds of a lot of wheat, amounting to $1275, I purchase with that sum a draft at 1 g % discount ; required the face of the draft. Ans. $1289.506+. — 2. What is the face of a draft for 15 days, which may be pur- chased for $1050, at 1^% premium? Ans. $1037.549+. -^3. What is the face of a bill on Boston for 60 days, at -^ % premium, which may be bought for $3000 ? y FOREIGN EXCHANGE. 34:9* The method of computing foreign exchange is similar to that of computing inland exchange, except that the currency of one country must be reduced to that of another. 3cS0* The value of 1£ sterling, which is the English sov- ereign, compared with the old United States coin, equals $4.44|. But Congress has from time to time reduced the weight and purity of United States coins, making their value as metals less than their value as coins, that they might not be used for transportation or 234 PERCENTAGE. the arts, and has established the legal value of the pound sterling at $4.84. The intrinsic value of the pound is $4,861. The com- mercial value varies from $4.83 to $4.87, as it is in greater or less demand. S5t, Exchange, however, is reckoned upon the old or nom- inal value of the pound ($4.44|), and the present value is said to be at about 9 % premium. Thus, when exchange on England is quoted at 10 or 11% premium, it is really only at about 1 or 25^ premium upon the real value. 309. Examples in Heduction of Currency and Ex- change. 1. What is the nominal value of £250 sterling expressed in United States money? $4.44| = rV. ^^/-4^ = $1111.11^, Ans. 2. What is the United States legal value of the above ? Ans. $1210. 3. Reduce 1£ sterling to Federal money at 9^% advance upon the nominal value. Ans. $4.86|. 4. Reduce 40£. 10s. to Federal money, at 9^ % advance. (See Art. 265, Note.) 5. What will be the value in Federal money of 84£. 19s. ll|d. at 10% advance? 6. What is the cost in New York of a bill on Liverpool for 470£. 13s. 9d., at 9|% premium? 7. A gas company purchased Newcastle coal in England to the amount of 1000£. 15s. 7f d. ; exchange being 8^% premium, what will this be worth in United States money ? 8. A dealer in flour shipped to London 3000 barrels of flour, which cost $4.20 a barrel in Baltimore ; the flour was sold in London at 1£. Gs. 6d. per barrel, exchange being 10% advance; what was the gain, without regard to expenses ? 9* How much Federal money will pay for 3 T. 15 cwt. 2qr. 1 lb. of iron, at 7£. 10s. 9d. sterling per ton, when the premium is 9^%? -<10! A cotton broker sent to Manchester, England, 50 bales of EXCHANGE. 235 cotton averaging 460 pounds each; the cotton was sold at lid. per pound. What was the amount of sales in United States money, premium being lOf %, and what was the broker's com- aaission at l^fo upon the sales? jf ,9 / i^ 11. Reduce $200 to sterling money when exchange on Eng- land is 10% advance. 200 x 9 x 100 _ 40£. I8s. 2^yd., Ans, /'' 40X110 11? 12. Reduce $575 to sterling money at 9f % advance. 13. When exchange is 9 5^ premium, what is the value of $6874.40 ? 14? What will a merchant gain by buying 4000 bushels of corn in Albany at 65 cents per bushel, paying for shipping and other expenses 30 cents per bushel, and selling it in Liverpool, England, at 4s. 9d.per bushel, when exchange is 10% premium? 15! A merchant in 1864 shipped to Liverpool 5000 pounds of '' butter, which cost him in New York 35 cents per pound; he"'*"'' paid 4fo for freight and other expenses, and sold it in Liverpool ^ for lid. per pound. Exchange being 120% premium, did he '^ gain or lose, and how much ? v 16. What is the cost of a set of exchange on Paris, for 6OOO ^^ francs, exchange being 6| francs per dollar ? 17. What is the cost of a set of exchange on Paris for 4500 ^ j francs at 5% premium, the value of 1 franc being 18| cents ?^^^ 'K f ^^ '/■ 18. Mr. James Payne, an American gentleman, while trav- elling in England received the following draft; what was the cost of the draft in America, the premium being 41 % in favor of England ? £127 Boston, Au^. 23, 1863. At sight of this, our jirst Bill of Exchange (second and third of same tenor and date unpaid), pay to the order of James Payne, in Manchester, one hundred and twenty-seven pounds sterling^ value received, and charge the same to our account. J. E. Thayer & Bro. ^) To Messrs. Calmont Bros. & Co, Bond Street, London, j j ^v 1^ ^£ 3« Questions in Heview. "What is Percentacje * From what is per cent, derived ? In what four ways represented? Represent |^ decimally. How is a per cent, reduced to lowest terms ? How is a common fraction reduced to a per cent. ? What is the complement of any rate per cent. ? Reduce | to a per cent. ; represent it decimally ; find its comple- ment, and reduce that to its lowest terms. How do you find any required per cent, of a number ? Name the applications of percentage in this book. (See Contents.] Upon what is the percentage of Profit or Loss reckoned ? If goods cost 24 cents, for what must they sell to gain 8^^ ? to lose 1 6f ^ ? What per cent, would be gained or lost by selling the above at 30 cents ? at 21 cents ? If 24 cents is 20^ less than the value of goods, what is the value ? if 24 cents is 33^^ more than value ? If 18 cents is 10^ less than cost, for what would you sell to gain 10^ t to lose 25^ ? If 10^ of what you receive for goods is gain, what is your gain per cent. ? What is Interest ? principal ? amount ? legal rate ? usury ? Usual rate in United States? Rule for finding interest on $1 a1 6^ ? on any sum ? At 6% in what time will a sum double ? will the interest equal ^ of the principal? J? ^? 1? i? i? ^? tV? l^? ^? ^V? i^ ? ToW? At 6^, what part of the principal is the interest for 1 mo. ? foi 3 mo. ? 5 mo. ? 6f mo. ? 11 mo. ? 13 mo. 10 d. ? 1 y. 4 mo. 20 d. ? 1 y. 8 mo. ? 5 y. 6 mo. 20 d. ? What is the interest on $1 for 3 y. 4 m. ? 6 d. ? 3 d. ? 1 d. ? 20 d.J 33 J mo. ? How obtain interest at any rate besides 6^ ? How obtain interest on pounds, shillings, &c. ? What is the rule for reducing shillings, pence, and farthings, to decimal of l£ by inspection ? What is the value of Is. in decimal of £1? 3s.? OS.? Iqr.? 12 qr.? 6d. ? 9d.? Name the first month in the year ; the fourth ; the seventh ; the tenth ; third ; twelfth ; ninth ; fifth ; eighth. How many months and days forward from June 7th to the 2d of each of the above ? from November 15th ? from March 28th ? What are Partial Payments ? AVTiat is the legal rule fcr partial payments ? How is the record of payments kept ? What are the pay r.UESTIONS IN REVIEW. 237 jnts called? Suppose an endorsement will not cancel the interest? ; lie in common nse when the note is paid within one year ? Rule for nual interest ? What is Compound Interest ? How often may interest be com- •unded ? For how many periods of time will interest be compounded 2 y. 9 mo., if it is compounded semi-annually ? quarterly ? monthly ? What three factors are employed to produce interest ? The inter- t, principal, and rate being known, give the rule to find time. The terest, principal, and time being known, give the rule to find rate. The terest, rate, and time being known, give the rule to find principal, /"hat is the dividend in each case ? To find time, for what time do )u take interest for a divisor ? to find rate, at what rate ? to find •incipal, on what principal ? Amount, rate, and time being known, ve rule to find principal. What does Present Worth embrace ? What is discount ? Give de for present w^orth. How find discount ? How prove the work ? What is a Promissory Note ? What the face of a note ? What bank discount ? What are days of grace ? avails of a note ? Which is the larger, true or bank present worth? true or bank iscount ? Describe the process of getting a note discounted at a bank. What is endorsing a note ? Rule for finding the face of a note, which, discounted at a bank, ill yield a certain sum ? What is Commission ? Brokerage ? What are Stocks ? When are ocks above par ? below par ? at an advance ? discount ? premium ? Upon what is the per cent, of commission or brokerage estimated ? What is the rule for finding what sum is to be laid out when a re- ittance is made which contains that sum together with the commis- on ? How obtain commission ? What is Insurance ? policy ? premium ? W^hat are underwriters ? What is the rule for finding the policy which will cover a certain im and the premium ? What is Equation of Payments? the interest rule? Pvule for juating an account ? What are Taxes ? polls ? real estate ? personal property ? assessors ? :ow find tax on $1 ? What are Custom-houses ? Duties ? What is a specific duty ? an i valorum duty ? leakage and breakage ? tare ? tonnage of vessai^ ? f08s weight ? net weight ? 238 PERCENTAGE. j What is a Draft ? Who is the drawer ? the drawee ? the payee i the holder ? What is the endorsement of a draft ? the acceptance ? What is a Bill of Exchange ? What is Inland Exchange ? For- eign Exchange ? When a bill negotiable ? when at a premium ? when at a discount ? Define course of exchange ; a set of exchange. Give the rule for finding the value of a draft ; for finding the face of a draft. What is the nominal value of 1£ sterling? the legal value? the intrinsic value ? Upon what is Sterling Exchange reckoned ? 3^4. Miscellaneous Examples. 1. Reduce -j-V to a per cent. 2. Represent Ig^'g^ decimally. ^ 3. Reduce 106^^ to its lowest terms. ^' 4. Find the complement of 84^%. 5. What is 124% of 5 T. 3 cwt.? 6. What is the amount at 6%, simple interest, of |o8.75, from Aug. 5 to Nov. 10 ? 7. What is the amount of $380.25, at 6% compound interest, for 2 yrs. 5 mo. ? 8. What is the simple interest, at 5%, of 10£. 8s. 6d., for 4 yrs. 2 mo. ? 9. What is the compound interest of the above at the same rate and for the same time ? 10. $2500. Chicago, April 4, 1862. In three months from date, I promise to pay John Peirce, or bearer, twenty-five hundred dollars, with interest after at 6^, value received. P. T. Ivison. The above was endorsed as follows : $1010 March 28, 1863; $100 Aug. 10, 1864; $1000 Jan. 1, 1865. What was still due? 11. At what per cent, must $450 be on interest 9 months to gain $13.50? 12. I received $7.35 for the use of $150 a certain time at 7%. Required the time. MISCELLANEOUS EXAMPLES. 239 13. Lent a certain sum for 1 y. 6 mo. at 5% ; the interest be- ing |>9.40, What was the sum? j ? /, ^ -■) 14. What principal will amount to ^63.25 in 1 y. 3 mo. at 8%? 15. $150.25. WiNTiiROP, Jan. 5, 1860. On the fifteenth of May, 1860, 1 promise to pay to the order of B. F. Tweed one hundred fifty ^J'q dollars, with interest after, at 6%, value received. D. D. Daniels. If the holder of the above have it discounted at a bank Feb. 15, 1860, what will he receive ? 1 6. "What will be the true present worth of the above at its date? /-^ f^* "x non-resident who owned a house in town valued at $2000 ? 35. E.f?4rce 750£. 10s. 4d. sterling to United States currency, exchange at 17% advance upon the nominal value, i _^ !^^"\\ J 36. A debtor owes $200, ^ due in 2 months, ^ in 3 months, and the remainder in 5 months ; what is the equated time for paying the whole ? ; 37. If ^ of a sum of money be due in 2 months, |- in 4 months, ^ in 3 months, and the remainder in 4 months, in what time should the whole be paid ? 38. What is the average time for paying for $200 worth of goods purchased May 17, 1859, on 4 months' credit; $500 worth, purchased June 18, 1859, on 60 days' credit ; and $300 Avorth, purchased June 19, 1859, on 90 days' credit? . 89. There is a note, dated July 1, on 60 days' credit, for $200. EXAMPLES IN PllOm^ AND LOSS. 24!. July 20 there is paid $50; Aug. 19, $60; and Aug, 21, $10. When should the balance be paid ? 40* Thomas Swain owes Wm. C. Chapin $500,18, due Jan, 7, 1863, and $207.06, due April 4, 1863; Mr. Chapin owes Mr. Swain $800, due Feb. S, 1863. When should the balance be paid? 4lt What is the cash value of the above March 25, 1863 ? 42f What is the date of a note which must be given to settle the following, allowing 4 months' credit on each of the merckan" dise items? John F, Stone in % with Thomas Emerson's Sons. Dr. Cr. 1863. Sept. 15. Dec. 1. 1864. Feb. 10. Mar. 1. Jan. 9. " 25. Apr. 1. To 200 bbk. Apples, (a) $3.25, 100 *' Flour, « 3.48, " 35,000 Bricks, « Cash, . . ^ By Merchandise, , do., " Cash, . . . 6.00, per M. 650 300 00 00 510 476 265 Do^S. 3^^* Miscellaneous Examples in 3?rofit^and Note. — Younger pupils can omit this article till they review. 1. By selling goods at 6 cents 3 mills per yard, I lose 30% ; what did they cost ? 2. If I lose 10% bj selling goods at 18 cents per yard, for what should they have been sold to gain 20 % ? 18X100 = 20 cts.i the cost : 20 X I'-iO zzz 24 cts. Ans, JW 100 3! By selling a lot of iron at 12^% below cost, I received $14.75 less than I should have received if I had sold it at 12^% above cost; what did it cost? what should it have sold for to gain 12^% ? 4. A merchant bought 5 cwt, 1 qr. of coifcc for $03 ; for wh.iit must he sell it per lb. to gain 16§% ? 5. For Avhat must hay be sold per ton to gain 13^-%, if by selling at $16, 33^% be gained? Note.— $16 -5- 1.33^ = |5l2; 1,13^ of $12 = $13,62, Anu 16 242 SCPERCENTAGE. /! 6. If 12^% bo gained by selling ladies' slippers, at $9 per doze^i pairs, for what should they have been sold per pair to gain oD%? Ans. 90 cents. 7. 13% is lost by selling a lot of land for $783 ; what would it have brought if it had been sold at a loss of 8^% ? A71S. S825. 8. 50% of a certain number exceeds 35% of it by $13.70 ; what is the number? Ans, $91.33^. 9. A person takes a note on 2 months' credit for $110 in pay- ment for a watch ; on getting the note discounted at a bank, he iinds that he has lost 40 % on the first cost of the watch. Re- quired the cost ? Ans. $181.40f. 10. A broker purchases a lot of stocks at an average of 9% below par, and sells them at an average of 7f % above par, and makes $300 ; what was the par value of the stocks ? / ( ; 11. If, by selling goods at 60 cents per lb., 20% is gained, what % would have been gained by selling them at 75 cents per lb. ? Ans. 50%. 12. If 10% is lost by selling boards at $7.20 per M., what % would be gained by selling them at 90 cents per C. .'* 1 13. Sold boots at $3.55 per pair, and thereby lost 29% ; what % would have been lost by selling them at $57.50 per dozen pairs ? 14. A dealer has 18 barrels of sound apples remaining in a lot of which 10% have decayed; if his lot cof.t him $1.50 per bbl.. would he gain or los3 on the lot, and what % by selling the re- mainder at $1.75 per bbl. ? Ans. 5 % gain. . 15. Sold 4 ploughs at $24 each ; on 2 of them I made 20%, and on 2 I lost 20%; what did I gain or lose on the whole ? Ans. Lost $4.00. 16. 20% of a lot of barley, originally 5000 bushels, was de- stroyed by fire, the cost having been $1^ per bushel ; what per cent, will be gained on the lot by selling the remainder at $2 per bushel ? 17. By losing 3 cents on a pound, I get 87|-% of the cost of butter ; if I had lost 4 cents a pound, what % should I have receivecl.^ J[w5. 83]%. EXAMPLES IN PROFIT AND LOSS. 24S 18. By having 5 pupils absent from school my attendance is 93|%; if my attendance had been 95^, hoAV many pupils must have been absent ? Ajis. 4 pupils. 19. What will be the %of gain on the cost of 174 shares of Roxbury Gas Co.'s stock, nominal value of shares being $87.50, if it be bought at 15% below par, and sold at 19^% above par? ^ > 20. If I buy coal at $4.12 per ton on 6 months' credit, for what must I sell it immediately to gain 10^^? Ans. $4.40. 21. If I pay $3.90 cash, for what must I sell it on 4 months to gain 20% ? 22. If, by .>clling wood at 75 cts. per cd. ft., 6|% be lost, for what should it have been sold per cord to gain 34%? Ans. $6,624. 23. I sell ^ of a lot of goods for $9, and thereby lose 25%; for what must I sell the remainder to make 8^% on the whole ? Ans. $30. 24. A grocer bought 100 gallons of oil, at $1 per gallon ; he mixed with it 50 gallons that cost $1.60 per gallon, then sold the /7^ whole at the rate of $1.44 per gallon ; did he gain or lose, and ^L what % ? § : ^ 25. Suppose the above was sold on a credit of 6 months, what was the % of gain? 26. Suppose the oil to have been bought on 4 months, and sold for cash ; what % was gained ? ^S^X ~^ 27. For what should he sell the above per gallon to make 2a %, if he bought for cash and waited 10 months for his pay f " ; 28. How much water must be mixed with a barrel of ink (31 galls.), which cost $34.10, that it may be sold at $1.10 a gallon, and a profit of 25% be realized by it? Aiis. 7| gall, 29. What water would be required in the above, if the cost had been $25, the profit 20 % , and the selling price $.75 per gallon ? 30. If 20% of what I receive for an article is gain, what is the gain % ? Note. — If 20j^ is gain, 80^ is cost; the gain then ^ ^ of the cost, which equals 25^, Ans. 31. If 25% of what I receive is gain, what is the gain % ?, ^o 244 PERCENTAGE. 32. When two sovereigns are sold for 124.20, what is the premium on gold, estimating the sovereign at its legal value ? j Jy/} Bo. What amount of current money must a broker give me for 120 in gold, when the premium on gold is 178% ? TT,P ^ 34. A man hands a broker $40 in currency, saying, " Give me the value of this in gold ; " gold is " quoted " at $2.64 ; how many dollars did the broker return, and how much change in currency ? 35. Suppose the broker paid 20% premium for this gold, what did he gain upon the $15, no allowance for interest ? 4 S/^'^ 36. What % did he gain on the money he expended to pur- chase the gold ? I ^ ' ^- 37. A man about to travel in Canada bought $50 in gold, when gold was at a premium of 150% ; what amount of current money did he pay for it ? / 2 0\ ^ 38. When gold is at a premium of 50%, what is the depreci- ation in the currency ? 39. A merchant imports from Hamburg a bale of cloth, con- taining 12 pieces, 40 yards each ; the cloth, with charges there, cost him $480 ; he pays a duty here of 35 cts. per yd., freight $28.50, and other charges $7.11 ; at what must he sell the cloth per yd. to gain 10 % above all charges ? / M ^-(^ ^ -j^ 40.* Vinton & Morris imported a lot of W. I. sugar, for which they paid $15272 in Havana; their expenses amounted to 5% of the first cost; 10 months after, they sold it so as to gain $5S12.46 above all expenses, including interest at 6%; what % Aid they gain ? f ' - \^ 41. An importer received from Holland 75 gross quart bottles of beer, invoiced at 10 cts. per bottle ; a duty of 30 cts. per gal. was paid at the custom house after 10% had been deducted for breakage ; if all the bottles be found sound, what % will be gained by selling the beer at 25 cents per bottle ? 3^6. General Revievt, No. 7. 1. What is } per cent, of $56.49 ?.^ ' 2. What is the amount of $84.20 for 3 yrs. 5 mo. 12 ds., at 5 per cent. ? 1 '^ . ( ' GENERAL REVIEV/. 245 3. What is the interest of 24 £. 7 s. 6 d. for 2 yrs. 4 mo. ? 4. Oct. 12, 1861, gave my note on demand, with interest, for $480. Feb. 6, 1862, paid $120. What remained due Aug. 24, 1862? 5. I held a note for $500, which bore interest from May 10, 1859. Sept. 16, 1860, received $140; July 28, 1862, received $50. What remained due Sept. 4, 1862 ? ^1),)^^ 6. If I pay $45 interest for the use of $500 for 3 years, what is the rate per cent ? 3 7. How long must $204 be on interest at 7 per cent, to amount to $217.09? I I r^b 8. What principal will gain $9.20 in 4 mo. 18 ds., at 4 per cent. ? 9. What sum, at 7 per cent., will amount to $221,075 in 3 yrs. 4 m.? 10. What is the compound interest of $600 for 1 yr. 4 mo., interest payable semi-annually ? / V , w 11. What is the present worth of a note for $488.50, due in 2 yrs. 5 mo. 15 ds., at 9 per cent.? 12. What is the discount of $105.71, due 4 yrs. hence ? 13. What commission must be paid for collecting $17380, at 3^ per cent. ? 14. What amount of stock can be bought for $9682, and allow \ per cent, brokerage? 15. What is the value of 20 shares bank stock, at 8^ per cent. discount, the par value of each share being $150 ? ^ / ^ S^f 16. What sum will be received from a bank for a note of 1260, payable in 4 months ? -; ^s ^\ \ I \ ""^ 17. What is the bank discount on $320 for 90 days ? 18. What is the face of a note which yields $112,803, when discounted at a bank for 60 days ? \\^\\ 1 9. A house, valued at $4750, is insured at J of 1 per cent. ; what is the premium ? j /; i. J lyO 20. What is the duty, at l5 per cent, ad valorem, on 20 bags of coffee, each containing 115 lbs., valued at 42 cts. per lb.? y//i/ ?46 EATIO. RATIO. Scir* Hatio is the relation which one number bears to a lother number of the same kind. Ratios are of two kinds, Arithmetical and Geometrical. 3^80 Arithmetical Ratio is ratio of numbers with respect to their difference ; as 6 — 4=2. GEOMETRICAL RATIO. 359. Geometrical Ratio is ratio of numbers with respect to their quotient ; as 2 : 4 = 4, read 2 is to 4, or the ratio of 2 to 4 =: J ; 6:3 = 2, read 6 is to 3, or the ratio of 6 to 3 = 2. 300. The first term of a ratio is called the Antecedent, the second, the Consequent ; both together are called a Coup- let. What is the antecedent in the first illustration in Article 359 ? the consequent in the second ? the ratio in the first ? the consequent in the first? the ratio in the second ? 301. When the terms of a ratio are equal, the ratio is one of equality; when the antecedent is greater than the conse- quent, it is a ratio of greater inequality ; when the antecedent is less than the consequent, it is a ratio of less inequality. 36^« It will be readily seen that ratios, being expressions for division, are similar to fractions. They can at any time be ■ written in a fractional form, the antecedent taking the place of the numerator, and the consequent that of the denominator. The principles applicable to fractions apply also to ratio. Hence, Multiplying the antecedent, ) , . ,. 7. .7. ,7 , V multiplies the ratio. or dividing the consequent, ) ^ Dividing the antecedent, t 7. . , ^ divides the ratio. or multiplying the consequent, ) Multiplying or dividing both ter of a ratio by the same number y Multiplying or dividing both terms ) . 7 . , r does not alter its value. GEOMETRICAL RATIO. 247 363. Ratios, like fractions, may be simple, complex, or compoimd. A ratio is sim.ple when each term is a simple number; it is complex when either term contains a fraction ; it is compound when it is the indicated product of two or more ratios. Simple Ratio. Complex Katie. Compound Eatio. 2 24 2:8. j--f, 2X3:5X5. 364:, Exercises. 1. Write the ratio of 2 to 3; of 7 to 10 ; of f to f ; of 2 X 7 to 5 X 4. 2. Multiply the ratio o : 4 by 2. 3. Divide the same by 2. 4. Reduce the ratio 6 : 6 to lower terms. 5. Write any ratio of equality ; of greater inequality ; of less inequality. 5 300. III. Ex. Reduce f : — to a simple ratio. ^^ Opekatiox. -U . Multiplying each term of the ratio | : J^^- by 3 X 7, we have : — r=14:45, Ans. Hence, jo A To reduce a complex ratio to a simple one : Reduce each term to its simplest for m^ then multiply each b^ the least common multi- ple of the denominators, and cancel. Reduce to simple ratios, 1. 2x = l. 2. ^ = 8. 3. 3X5:4X8. 4. 5x4:3Xf 5. |X5:|X1| 248 PKOPORTION. PROPORTION. 360:) Proportion is an expression of equality between two ratios ; thus, 2:3 = 4:6, read 2 is to three as 4 is to 6 ; that is, 2 is the same part of 3 that 4 is of 6. 2 is f of 3, and 4 is § of 6. 36T, The first and fourth terms of a proportion are called the extremes, and the second and third are called the means. The first ratio is called the first couplet, and the second ratio the .second couplet. Read the following proportions: — 1. 5 : 10 =z S : 6. a 8:2 = 12:3 2. ^:lz=:5:15. 4. 06:7 = 8:1 Name the extremes of the first proportion; the means of the second; the antecedents of the third ; the consequents of the fourth j the sec- ond couplet of the first proportion. 308. Inverse Proportion. Four terms are directly proportional when the first is to the second as the third is to the fourth. They are inversely proportional when the first is to the second as the fourth is to the third, or when one ratio is direct and the other inverse. Thus, the amount of work done in any given time is directly proportional to the men employed ; i. e.> the more men, the more work ; but the? time occupied in doing a certain work is inversely proportional to the men employed ; i^ e., the more men, the less time. 30O« A compound proportion is an equality between a compound ratio and a simple ratio, or between two compound ratios. STO* Three terms are in proportion when the first is to the second as the second is to the third. The second term is called a mean proportional between the other two ; thus, in the pro- portion, 3:6=:6:12, 6isa mean pix)portional between 3 and 12. 3TI* The performance of arithmetical examples by pro portion depends upon the follow hig important principle: — SIMPLE PROPORTION. 249 In every proportion the product of the means equals the product of the extremes. Illustration. Writing the given proportion in a frae- Q . o 4-6 tionai form, we have f =: |. Muhiplying 2 A each fraction by the product of the de- ^ ^ . nominators, and cancelling, we have 2 X ^ X 3 X6 __ 4X3X0 6 z= 4 X 3. But 2 and 6 are the extremes, 3 and 4 and 3 the means ; hence the product 2X6 = 4X3 of the extremes equals that of the means. 373, From the abovo, it follows that whenever an extreme in a proportion is wanting, it can be found hy dividing the product of the 7neans by the given extreme ; and whenever a mean is want- ing, it may he found hy dividing the product of the extremes hy the given mean. Supply the terms wanting in the following proportions: — 1. 5:1 — 50:? 3. 9:7rr:?:5G. 2. 8 : ? = 3 : 10. 4. ? : 2 =: 15 : 5. 37S. In the proportion 2 : 4 == 4 : 8, 42 z= 2 X ^ -.4=:= V2 X 8 (Arts. 91, 92) ; hence a mean proportional between twd nmnbers equals the square root of their product. Supply the mean proportionals between the following numbers and w^rlte the proportions : — I 7. 2 and 24^. 6. 2 and 18. I 8. 20 and 5. 9. 3 and 27. 10. 16 and 4. ANALYSIS AND PROPORTION. ST^Ir, III. Ex., I. If 15 boxes of oranges cost $G0, what will 17 boxes cost? Opkuation by Analysis. If lo boxes cost $60, 1 box will cost ^ 4 of $60, and 17 boxes will cost 17 X t\ of ^60 X 17 ^60. Cancelling and multiplying, the re- X^ _ $>ow, n . ^^^^ .^ ^^^^ Operation by Pkopoutiox. $60, the price of 15 boxes, must bear the 15 : 17 ==: ^60 : Ans. same relation to the price of 17 boxes that 4 15 bears to 17. We have then three terms 17 X ^00 __ ^gg ^^^^ of a proportion (15 : 17 = $60 : ), and can 1^ ' find the fourth by multiplying the second and thu'd together, and dividing the product by the first. Hence wo derive the following 4 250 PROPOUTION. Et'LE FOR Simple Proportion. Mahe the mrniher that (s of the same hind as the required answer the third term. If the answer should he greater than the third term, make the larger of the other two numbers, upon which the answer depends, the second term, and the smaller the first. If it should he less, mahe the smaller number the second term, and the larger the first. Multiply the means together, and divide their product hy the first extreme. Nqte. — Analysis is the more natural and philosophical method of solving arithmetical questions ; but the principles of Proportion are applicable to certain classes of examples. It is recommended to the pupil to perform the following examples in both ways. He should, at Jeast, perform a sufficient number of them by Proportion to fix the method in his mind. Examples. 1. If 2 men build 17 rods of wall in a week, how many rod:* will 100 men build in the same time ? ,>: o ^^ ^' *"'^ We make 17 rods, which is of the same deno;t=iaation as the required answer, the third term. As 100 men will buiitl more wall than 2 men, we make 100 the second term, and 2 the first term, and the statement is, 2 : 100 — 17 : 850. Ans. 850 rods. 2. If 9 lbs. of lead make 150 bullets, how many bullets can be made from 105 Ibs^? Ans. 1,750 bullets. ' 3. If 65 pairs of boots can be made from 75 lbs. of calfskin, how many pairs can be made from 850 lbs. ? ' Ans. 73 6§ prs. 4. How many tons of hay can be made from 750 acres of land, if 13 tons can be made from 3 acres ? Ans. 3250 tons. 5. If $2000000 will support an army of 500000 men a day, • how many men can be kept for $400000 ? Ans. 100,000 men. 6. If $500 purchase 200 hats, how many hats can be pur- chased for $87^^ ? Ans. 35 hats. 7. If $800 yield %i)(d of interest in a certain time, what will $390 yield at the same rate? Ans. $27.30. . 8. If 16 horses eat a certain quantity of hay in 13 weeks, how long would the same quantity last 24 horses ? Ans. 8f weeks. 9. What time would be required for 5 men to mow an acre of land, if 2 men can mow it in 1 J- days of 10 hours in length ? Aris. 6 hours. ANALYSIS AND SIMPLE PIIOPORTIOX. 2ol 10. If o3v) bushels of plaster were sufficient for the dressing of 3^ acres of land, what would be required for 17^ acres ? Ans. 2500 bu. 11. If 95 acres of land were mowed by 20 men in 2 days of 12 hours each, how much time would be required for 3 mowing machines to do the same work (1 machine being equal to 4: men) ? Ans, 3 d. 4 h. 12. If a Graham loaf weighs 1 lb. 2 oz, when flour is worth $7 J- a bbh, what should it weigh, selling at the same price, when flour is $6 per bbL ? Ans. 1 lb. 6 J oz. 13', If 400 lbs, of coal are required to run an engine 12 hours, what number of tons will be required to run three similar en- gines for 30 days constantly ? ) 1 >- /: 14. If it requires 13 days of 9 hours each to do a piece of work, liow many days of 14 hours each will be required to do the same work ? Ans, 8 y^j. 15. If soda-crackers can be sold at 10 cents a pound wheu flour is worth $6.50 per bbl,, for what can they be sold when flour is worth $9.75 per bbL ? / h^ '' 16. If it requires 30 men to do a piece of work when they work 1 1 hours a day, what number will be required when they work 15 hours a day? > .C 17. If 30 men, working 11 hours a day, can do a piece of work in a certain time, how many more men must be employed, when it is half done, to finish it in. the same number of days, working 10 hours a day? Ans. 3 more. 18. A pole 10 ft. long casts a shadow of 7 ft. 1 in. at 8 o'clock ; what is the height of a flagstaff that casts a shadow 58 feet at the same time of day? Ans. 81|^ ft. 19. If my friend lends me, $7000 for 15 days, for what time should I lend him $4500 to requite the favor? Ans. 23^ d, 20. If my friend lend me money for 4 months when interest is 10 per cent., for what time should I lend him the same sum when interest is 7 per cent. ? jj* -^ 21. A sail vessel has 10 hodrs the start of a steamer, and sails at the rate of 7 miles an hour, while the steamer sails 16 mile^ an hour ; when will the steamer overtake \lie sail vessel ? Ans. 7| h. 252 PROPORTION. 22. A deer, 150 rods before a hound, runs 30 rods a minute; the hound follows at the rate of 42 rods a minute ; in what time will the deer be overtaken ? / % ', j l^^Q \\ ] \ ( I l ^ 23. Two armies are on opposite sides of a river, one being 500 miles east and the other 250 miles west of it, and marching jowards each other, the first at the rate of 15 and the other of 18 miles in a day ; in what number of daj's will they meet, and where? ^?is. IGf d. ; 50 m. east. 24. If 2 lbs. 5 oz. of wool make 1 yd, of cloth 32 inches wide, how much will make a yard 1^ yards wide of the same quality ? Ans. 3 lb. 47jV oz. 25. How much cloth that is | yd. wide will cover 24 tables 6 ft. long and 3 ft. wide ? Ans. 64 yds. 26. If it requires 10 yards of cloth that is 1 1 yds. wide to make a garment, hovy much will be required of that which is 1 1 yds, wide ? / 27. How many yards of cambric 24 inches wide will be re- quired to line 14^ yards of sillc which is 22 inches wide ? . J^ 28. How long must a piece of land be to contain 3 acres, if it* is 4 rods wide ? 29. If 400 bushels of potatoes were bought for $350.90, and sold for $425.50, what would be gained on 25 bushels ? 30. A man failing in trade owes $7,865 ; his property is sold for $4385.70 ; what will he be able to pay to a creditor to whom he owes $1500 ? 31. If it costs $30 to paint the outside of a house 40 ft. by 30 ■^t. and 25 ft. high, what will it cost to paint one 50 ft. by 40, of the same height? A7is. |38f. 32. If a building 13 ft. high casts a shadow of 4 ft., what is the length of a shadow cast by a tree 346| ft. high at the same time? , , .' 33. If $110 be paid for 3 T, it cwt. 20 lb. of hay, what will be the cost of 6 T. 3 cwt. 3 lbs, ? Ans. $78.59 ^^g^. 34. If a hind wheel, which is 8|- feet in pirciimfeTence, turns 800 times in a journey, how many times will the fqre wlicel which is 6^ feet in circumference, turn in the same journey? ANALYSIS AND SIMPLE PROPORTION. 2lB 85. The weight of a cubic foot of water being 62^ lbs., how many pounds' weight will a tank contain which is 2^ ft. square at the bottom, and 4 ft. high ? / ' > 2 ^^ 36. If 15 A. 2 R. 20 r. pasture 2 cows a certain time, what will be required to pasture 25 cows the same time ?y ^r , v. / 37. How many yds. of cloth can be bought for 17£. o s., if 10 s. were paid for 5 quarters, or 1 Ell English ? 38. What is the value of 7 cords 3 cord feet of wood at the rate of $18 for 2 cords 5 cord feet ? 39. If 5| cwt. of leather pay for ^ of an acre of land, how many pounds would be required to pay for 1^- sq. rods? Ans. 12-j3/^ lbs. 40. If 9 oz. 10 pwt. 3 gr. of gold be worth 8150, what will be the value of 7 lb. 5 oz. 5 pwt. ? /■ f ^ 1/ o\ / '^}'iJ 41. If 15|.lbs. of tea pay for 48 lbs. 12 oz. of butter, how many pounds of butter can be purchased with a box of tea which con- tains 42 1 lbs.? j " ,. . : ' :■ 42. If .80 Jy acres of land be worth 175.20, what is the value ^f .373^ acres ? Ans. $35,053+. Note. — Perform the following examples by analysis. 43. If a cow and a calf sell for $27, and the value of the calf is I that of the cow, what is the value of each ?^ 44. The sum of the ages of a father and son is 48 years, that of the son being | of the age of the father ; what is the age of each ? Jcr 45. By one pipe a cistern can be emptied in 2 hours ; by an- other it can be emptied in^ 3 hours : in what time can it be emp- tied if both are running? -^^^ f "^" '- i" Z" YJ/ - '' / 46. A certain cistern h^s .^ pipes ; t«e"fSst will 4mp£y it in 5 hours, the second in 4 hours, and the third in 10 hours ; in what time will all empty it ? | j^ . 47. A certain piece of work can be performed by A in 8 days, by B in 10 days, and by ^ in 1^ days; in what time can all d^^ it working together ? -5 V': ^. 48. In what time can £ and B do it together ? ^i i: 254 PROPORTION. 49. In -what time can A and C do it together ? 50. In what time can B and C do it together ? 51. A cistern has one pipe which will fill it in 5 J- hours, and another which will empty it in 33 h. ; in what time will it be tilled with both open ? Ans. 6f h. 52. If A and B can do a piece of work in 3 mo. 10 d., and A alone can do it in 5 mo., in what time will B do it alone ? Ans. 10 mo. 53. If A and B can dig a trench in 6| days, and B can do it alone in 10 days, in what time can A do it alone ? :; 2 ' s 54. If A and B can do it in 5^ days, and A, B, and C,jcan do it in o^ days, in what time can C do it alone ? 55. If A and B can do a piece of work in 4| days, and A and C can do it in 5 ^ days, and B and C in 6y\ days, in what time? will all do the work together, and in what time will each do it alone ? Ans, A, B, C, 3^^d. ; A, 8 d. ; B, 10 d. ; C, i6 d. 56. A man sold a load of coal, containing | T., at $.75 for a hundred lbs., and received pay in corn at $.871 per bushel ; h<;w many bushels did he receive ? 57. Purchased a number of pieces of goods, each contain- ing 22 yards, at $4 for 3 yards, and sold them at $5 for 2 yards, and gained $154 on the lot; how many pieces were pur- chased ? Ans. 6 pieces- 58. "If a third of 6 be 3, what will a fourth of 20 be?" ^p" For Dictation Exercises, see Key. ANALYSIS AND COMPOUND PROPORTION. 375» III. Ex. If 10 gas-burners consume 800 feet of gas in 3 hours, how many feet will 12 burners consume in 15 hours? Operation by Analysis. If 10 burners consume 800 ^ feet of gas in 3 hours, 1 burner 800 X X^ X ^0 __ . , will consume ^\ of 800 feet, and j0-y-0 - — 4b00 tt., Ans. ^^ burners will consume 12 X ^ J^ of 800 feet in the same time ; if this is consumed in 3 hours, in 1 hour there will be consumed ^ of 12 X iV ^^ ^^^ ^^^^' ^"^ ^ ^^ hours, 15 X i of 12 X 1^^ of 800 feet. ANALYSIS AND COMPOUND rROPOllTION. 255 Opeuation by Pkopoktiox. 800 feet being of the same kind 10 : 12 ) __ gQQ . j^^Q^ as the answer, we make it the 3 : 15 * third term. But the number of 6 ^ feet required depends upon two 800 yC ^^ X X^ _ ,Q^^ /. J conditions: 1st, the number of X0X^ ~ *' ' gas-burners, and, 2d, the number ^ of hours during which they are burning. We shall deal with each condition separately. As 12 burners will consume more gas than 10 burners, we make 12 the second term and 10 the first term ; and as they will burn more in 15 hours than in 3 hours, we make 15 the sec- ond term and 3 the first term. We find the fourth term, as in simple proportion, hij dividing the product of the means, 12 X 15 X 800, hy the product of the first extremes, 10 X 3. The process may sometimes be shortened by cancelling the terms as they stand in the proportion, remembering that the numbers which constitute the first terms are divisors, and those which constitute the second and third are multipliers. Thus, 2 '^^i ^.^1^:800 :4800 ft., Arts, From the above we derive the following Rule for performing Examples by Compound Pro- portion. Make the number that is of the same kind as the an- swer the third term. Take each two numbers that are of the same kind, and consider whether, depending upon them alone, the answer will be greater or less than the third term. Arrange them as in simple proportion. Divide the continued product of the second and third terms by the continued product of the first terms. Examples. 1. If $90 is paid for the labor of 20 men 6 days, what should be paid for 5 men 8 days ? Ans. $30. 2. If 85 tons of coal were required to run 6 engines 17 hours a day, what number would be required to run 25 engines 12 hours a day ? Ans. 250 tons. 3. If 120 rods of watl were laid by 72 men in 33 days of 14 256 PROPORTION. hours each, how many rods would be laid by 88 men, working 12 hours a day for 3^ days ? Ans. 13^ rds. 4. If the wages of 75 men for 84 days were $G8.75, for how many days could 90 men be employed for $41.25 ? Aiis. 42 d. 5. If the freight on 450 lbs. of merchandise is 30 cents for 26 miles, how many miles can 3 tons be carried for $4? Ans. 26 m. 6. If, when flour is $7.50 per bbl., a 3-cent loaf weighs 2 oz., what should a 12-cent loaf weigh when flour is |16? Ans. 3| oz. 7. If ^ loaf, which sells for 10 cents when wheat is $2 a bushel, weighs 1^ lbs., what is the price of wheat when a 6-cent loaf weighs 1| lbs.? Ans. $1.44. 8. If 500 lbs. of wool worth 42 cents a lb. is given for 75 yds. of cloth 1| yds. wide, how much wool worth 36 cents a lb. should be given for 27 yds. that is 1^ yds. wide? Ans. 193}^ lb. 9. If it costs $135.02 to carry 7 cwt. 2 qr. 15 lb. a distance of 64 miles, what will be the cost of carrying 1 1 cwt. 2 qr. a dis- tance of 15J- miles ? Ans. $49,157+. 10. If 25 men, in 9^ days of 10 hours each, build 200 rods of wall, how many rods would be built in 1 day of 12 hours by 1^ men? Aiis. 12 if rds. 11. If 11 men, in 20 days of 12 hours each, can build a wall 48 feet long, 8 feet high, and 3 feet thick, in how many days can 15 men, working 10 hours a day, build 440 ft. of wall 12 ft. high, and 4 ft. thick ? Ans. 322| d. 12. How many men will be required, working 12 hours a day for 250 days, to dig a ditch 750 ft. long, 4 ft. wide, and 3 ft. deep, if it requires 27 men, working 13 hours a day for 62 days, to dig a ditch 403 ft. long, 3 ft. wide, and 3 ft. deep? Ans. 18 men, 13. If 5 men, working 12 hours a day for 8 days, cut 44 loadi of wood, each 8 ft. long, 4 ft. wide, and 4 ft. 6 in. high, in how many hours would 16 men cut 49^ loads 8 ft. long, 4 ft. wide, and 5 ft. 6 in. high? A?is. 41i hours. 14. If 7 cases of boots can be made by 9 men laborino- 12 hours a day for 7 days, what length of time will be required for 3 men and 4 boys (2 boys being equal to a man and a half), laboring 11 hours a day, to make 33 cases? Ans. 54 d. ANALYSIS AND COMPOUND PROPORTION. 257 15r If 421 1- rods of fence which is 4 feet high be built by 9 men in 10 days of 9 hours each, how many rods of a fence 5 ft. high will be built by 8 men in 6 days of 8 hours each ? 16* What length of canal 27 ft. wide, 24 ft. deep, can be dug by 250 men in 100 days of 12 hours each, if 750 men in 3 months of 25 days each, working 11 hours a day, could dig 4 miles of a canal 30 ft. wide and 10 ft. deep ? Arts. 7 fur., 7 rods, 5 ft., 2| in. 17!" If the type for a book of 84 pages, 50 lines to a page, lines averaging 8 words, 1^- syllables to a word, 2^ letters to a syllable, was set by 2 men in 5 days of 12 hours each, how many pages of a book, each page containing 75 lines, averaging 5^ words each, 2 syllables to a word, 3 letters to a syllable, would be set by 7 men, laboring 8 hours a day for the working days of a week? Ans, 142y\ pages. 18^ If 12 men, in 2 months of 4 weeks each, working 6 days per week, 12 hours a day, can set the type for 12 books, of 600 pages each, 120 lines to a page, 20 words to a line, 10 letters to a word, in how many months of 4^ weeks each will 7 boys, working 4 days per week, 16 hours a day, set the type for 6 books, of 500 pages each, 150 lines to a page, lines averaging 24 words, 4^ letters to a word, each boy doing ^ of the work of a man? Ans. l^f^jj months, 19t If it requires 15 yds. of silk, f yd. wide, to line a cloak made of 12.25 yds. of cloth, 1 yd. 1-i qrs. wide, how many yards of silk, I yd. wide, will be required to line a cloak made of 8.33 J yds., 4^ qrs. wide? Ans. 4jYf J^s. 20r If 4 men dig a cellar 33.75 ft. long, 18 ft. wide, and 9.6 ft. deep, in 4.5 days of 11.25 hours each, in how many days of 11.7 hours will 15 men dig a cellar 15.2 yds. long, 7.8 yds. wide, and 10.8 ft. deep, the latter cellar being twice as hard to dig as the former ? ^^ For Dictation Exercises, see Key. 17 258 PROPORTION. PARTNERSHIP, OR PARTITIVE PROPORTION. S'^0. Partnership is the association of two or more persons for the transaction of business. The persons thus associated are called Partners. The profits and losses are generally shared by the partners in proportion to the stock each has in trade, and the time it is employed. III. Ex. A and B formed a partnership for one year. A put in ^800, and B $700. They gain $375. What is the share of each? Operation. If ,91500 gain $375, $1 will $800 + $700 = $1500, whole stock. j^ _l^ ^f ^375 ^^^ ^8(^0 ^Vo of $375 := $200, A's share. .^-n g^l^sOO X -,i^^ of $375 iVA of $375 z=z $175, B's share. ^ ^200, A's share, and $700 will gain 700 X i^ of $375 = $175, B's share. Hence the KuLE FOR Simple Partnership. Apportion the gain or loss among the partners according to their stock in trade. Examples. 1. A and B form a partnership, A putting in $4000, and b $3000 ; they gain $3000 ; what is the share of each ? Arts. A, $1714f ; B, $1285^ 2. E, S, and Z invest in trade as follows : E $500, S $800, and Z $300 ; they gain $375.75 ; what is the share of each ? Ans £,$117.42^3^; S, $187.87i ; Z, $70.45-j5g. 3. Two traders, A and B, shipped coal from Philadelphia to Boston. A had on board 350 tons, and B 900 tons. It became necessary to throw overboard 250 tons. What was the loss of each? Ans. A, 70 tons; B, 180 tons. 4. Four persons, H, I, K, and L, engaged in an adventure H furnished $275 ; I, $640 ; K, $330 ; L, as much as II and K. They gained $4675. What was the share of each ? Ans. H, $694.93^9^; I, $1617.29§f; K, $833.91ff ; L, ^1528.85^^7. 5. A bankrupt owes to Q $800, to R $350, and to X and Y $500 each; his whole property sells for $1584.80, of which SIMPLE PARTNERSHIP. ^59 $158.48 arc required to pay the expenses of the sale ; what will each person receive ? Ans, Q receives $530.72|f ; E, $232.19/^; X and Y, each, $331.70^^. 6. Gerry & Frost enter into a dry goods business. G puts in 250 pieces of cotton, 30^ yds. each, at 10 cts. ; 100 yds. broad, cloth, at $3.25 ; and $918.75 in cash. F puts in 40 pieces linen, 12 yds. each, at 28 cts. ; other goods to the amount of |500, and cash $5G5.60. On closing business, they find they have lost 40 % of their investment. What is the loss of each ? 7. Four persons, M, N, O, and P, engaged in partnership. M put in 700 bu. wheat, at $1.25; N, 1000 lbs. wool, at |.37^; O, 500 bbls. flour, at $6.50; P, 2000 bu. corn, at $.90. Their whole stock being destroyed by fire, the firm received insurance on the goods, $4200 ; what was the loss to each partner ? /-.8. Suppose the store occupied by the above persons, owned by A, B, and C, and valued at $80000, to have been insured for f of its value. What would be the loss to each partner by the fire, the store being entirely consumed, A owning ^, B ^, and C the remainder? Ans. A, $5000; B, $6666f ; C, $83331. 9. Banks & Ward traded in company, and gained $975. It was agreed that B should have $8 of the gain as often as W had $7. What was the share of each? Ans. B, $520 ; W, $455. 10. Divide $1500 among three persons so that their shares shall be in the proportion of 3, 4, and 5. 11. Five persons. A, B, &c., hired a pasture for $275. A put in 250 sheep, B 300 sheep, C 200 sheep, D 15 yoke of oxen, and E 10 yoke. One ox being equal to 10 sheep, what was due from each? Ans. A, $55; B, $66; C, $44; D, $66; E, $44. 12!" F, G, H, and I are in partnership as stock brokers. F furnishes $2000, G $2500, H $1500, and I 75 shares of railroad stock. They gain $3300, of which I receives $1100 ; what was his stock per share, and what is the gain of each of the other partners? Jws. F, $733^ ; G, S916§; H, $550; I, $ 40 a share. 13! Stewart & Mills traded in company for one year. Their gain was equal to 20% of their stock. S's share of the gain was 2 GO PROPORTION. $150, which was | of the whole gain; what was M's gain, and what the sum each invested ? V /^ / 14? Adams & Brown built a schooner. A. furnished 18000, and B. $1700 and 15000 ft. of lumber. Her freights for the first year were $1125, of which B.'s share was $225 ; what was the price of his lumber per thousand feet ? Ans. $20 per M. 15t Jones, Styles & Carpenter enter into partnership. J. puts in $750, S. $420, and C. 60 tons of coal. They gain $624, of which C. is to have J for conducting the business, the balance to be shared among the partners in proportion to their stock in trade. C. receives $390 ; what is his coal per ton, and what are the shares of the other partners ? S^^ For Dictation Exercises, see Key. I' COMPOUND PARTNERSHIP. STT. When stock in trade is employed for different periods of time, the partnership is called Compound Partnership. III. Ex. Three persons formed a partnership. A put in $170 for 9 mo., B $130 for 12 mo., and C $150 for 8 mo. They gained $286 ; what was the share of each? Operation. The use of $ 1 70 170 X 9 = 1530 l|ff X $286 = $102, A's share, for 9 mo.z=the use 130X 12=lo60 ^ff X $286 = $104, B's share, of $1530 for 1 mo.; 150 X 8 = 1200 ii^ X $286 = $80, C's share, the use of $130 for 4290 12 mo. =$1560 for 1 mo. ; the use of $150 for 8 mo. =: $1200 for 1 mo. The amount in trade Avas, therefore, equal to $1530 + $1560 + $1200,= $4290, for 1 mo. ; hence the gains should be as follows : A's, i||g- of $286 rrr $102; B's,l||^of$286 = $104; C's, iff «- of $286 =$80. Hence the BuLE FOR Compound Partnership. Multiply each part- ner's stock hy the time it is in trade, and apportion the gain or loss according to the products. COMPOUND PARTNERSHIP. 261 Examples. 1. A and B engaged in business, and gained $2008.25. A put in $4500 for 9 months, and B $5690 for 7 months. What was the gain of each ? Am. A, $1012.50 ; B, $996.75. ^ 2. A, B, C, and D work a mine in company. A furnishes $1400 for 3 years, B $500 for 5 years, C $1800 for 2 years, and D $700 for 4 years. At the end of 5 years they divide $2620 of profits ; what is the share of each ? Ans. A, $840 ; B, $500 ; C, $720 ; D, $560. 3. Webb, Clapp, and Calhoun form a partnership. Webb puts in $8500 for 7 months, Clapp $10000 for 4 months, and Calhoun $6750 for 9 months. They lose $2499.90. What is each partner's loss ? ^m. Webb, $928.20; Clapp, $624; Calhoun, $947.70. 4. Hooker, Brown, and Lear traded in company. H. put i $2500 for 10 months, B. $2300 for 11 months, and Lear con- ducts the business, which is considered equal to $2000 in trade, for 12 months. They gain $1486. What should each receive ? Ans. Hooker, $500; Brown, $506; Lear, $480. 5. Four persons, J, K, L, and M, loaned money as follows : J $1500 for 5 years, K $750 for 3 years, L $1700 for 2^ years, and M $950 for 4 years. They received of interest money $1246. Wliat was the share of each, and what the rate per cent. ? Ans. J, $525 ; K, $157^; L, $297^; M, $266; rate, 7%. 6. A, B, and C formed a copartnership. A furnished |- of the capital for 6 months, B ^ of the capital for 10 months, and C the balance for 12 months. The whole gain was $1560. What was the share of each ? Ans. A, $480 ; B, $600 ; C, $480. ■^ 7. Hooker & Brown were in business together for 3 years, and gained $5750. Hooker put in $2000 for the first year, and $1500 for the other two ; Brown put in $2500 for the first two years, and $1500 for the last year. What was the gain of each ? Ans. Hooker, $2500 ; Brown, $3250. 8. A and B received $857.50 for grading a road. A furnished 5 hands for 20 days, and 6 others for 15 days; B furnished 10 hands for 12 days, and 9 others for 20 days. What was the share of each contractor ? Ans. A, $332.50 ; B, $525. 2G2 PROPORTION. 9. Lincoln and Hurd hired a pasture, for which they paid 8117. Lincoln jut in 217 head of cattle for 20 days, 150 for 5 days, 189 for 10 days, and 500 for 7 days; Hurd put in 650 head for 6 days, 48 for 15 days, and 400 for 11 days. "What should each pay ? Ans. Lincoln, $62.88 ; Hurd, $54.12. 4 10. Jones and Avery engaged in business as brokers for the year 1862. Jan. 1, Jones advanced $8600 and Avery $1250; April 1, Tyler was admitted to the firm whh |1500; June 1, Childs was admitted with $1200; Sept. 1, Hewins with $1800; and, Nov. 1, Jenkins with $2550. The losses for the year were ^7560 ; what was the loss of each partner ? Jws. Jones, $3534y6j.; Avery, $1227f\ ; Tyler, $1104f-j ; Childs, $687y\ ; Hewins, $589yiy; Jenkins, $417y^^. "n lit Wallis and Winn engaged in trade. The former had in $900 from January 1 till April 1, when he withdrew $450 ; July 1 he added $600. The latter had in $2000 from Feb. 1 to Oct. 1, when he added $200 ; Nov. 1 he withdrew $800. The whole gain during the year was $2500 ; what was the share of each. Ans, Wallis, $825/ij59 ; Winn, $1674^3|. wl2f Ames & Rice ran a steamer for 3 years. Ames furnished t|3000 for the first 10 months, when he added $1000 more, and at the end of the second year $500 more. Rice put in $2500 for the first 18 months, when he put in $3500 more. At the end of the third year they found their loss to be $5565 ; what should each sustain ? 13! D, E, and F hired a pasture on the 20th of May for 5 months, paying $125 for its use. On that day D put in 200 sheep, E 150, and F 80; June 20, D put in 40, E 200, and F 275 ; July 20, D took out 100, E 75, and F put in 80 ; Sept. 20, D put in 25, and E and F took out 200 each. What should each pay ? 14. Weeks, Wyman & Wentworth engaged in business for 1 year. Jan. 1, each put in $4000 ; March 1, Weeks and Wyman put in $1500 each, and Wentworth withdrew $600 ; Aug. 1, Weeks put in $800, Wyman withdrew $300, and Wentworth put in $1000; Oct. 1, Weeks withdrew $400; Nov. 1, Wyman put in $650, and Wentworth put in $1500. At the end of the year they divided $3500 profits. What was the gain of each ? REVIEW. 263 \ ISr A, B, C, and D put $5700 in trade. A's money was in 8 months, and his gain was |160 ; B's was in 5 months, and his gain was $200; C's was in 2 months, and his gain was $18); D's was in 6 months, and his gain was $240. What stock did each have in? A?is, A, |600 ; B, $1200 ; C, $2700; D, $1200. 1^ For Dictation Exercises, see Key. ^'- S78. Questions for Review. Ratio. — AVhat is ratio ? what is arithmetical ratio ? geometrical ratio ? What is the first term of a ratio called ? the second term ? hoth terms when taken together ? What is a ratio of equality ? of greater inequality? of less inequality? Give an example. In what respect do ratios resemble fractions ? How, then, may ratios, at any time, be written ? How do you multiply a ratio ? how divide a ratio ? Suppose you multiply or divide both terms by the same number ? What is a simple ratio ? a complex ? a compound ratio ? How do you reduce a complex ratio to a simple one ? a compound ratio ? Write a simple ratio ; a complex ratio ; a compound ratio. PiiOPORTiON. — What is proportion ? Explain the proportion 2 : 4 = 7 : 14. What are the 1st and 4th terms called ? the 2d and 3d ? the 1st and 3dP the 2d and 4th? the 1st and 2d? What is inverse proportion? compound proportion? AVhat is a mean proportional between two numbers ? Upon what important principle does the solv-. ing of examples by proportion depend ? Write a proportion, and illustrate that principle. How can you find an extreme, when the othet three terms are given ? how a mean ? how a mean proportional be- tween two given numbers ? Give the rule for solving an example by simple proportion, and illustrate it by an example of your own. Per- form the same example by analysis. What is meant by analysis ? Give your rule for solving an example by compound proportion, and illus. trate it. In solving any example by proportion, the two terms of t\ ratio must be of the same kind; why? Partnership. — What is partnership ? Who are the partners ? How are ])rofits and losses usually shared ? What is simple partner- ship ? {Alls. It is partnership where persons enter into business for the same time.) How do you find each person's share of gain or loss in simple partnership ? AVhat is compound partnership ? How do you find the shares of gain or loss in compound partnership ? Why is Vartnership sometimes called partitive proportion ? 264 INVOLUTION. yiNVOLUTION. 379. Involution consists in raising a number to a required power. (Art. 89.) 380. The required power is indicated by a small figure, called the index or exponerit, placed at the right, and a little above the number. (Art. 90.) 38 !• The Jirst power of a number is the number itself. The second power or square of a number is obtained by using the number as a factor twice. The third power or the cube results from using the number as a factor three times, and so on. Note. — The most important applications of Involution are in the use of the second and third powers. 38S* Any power may be obtained by the following Rule. Einploy the given number as a factor as many times as there are units in the exponent of the required power. Examples. 1. Find the squares of the integers from 1 to 25 inclusive, Qnd commit them to memory.* /Numbers, 1, 2, 3, 4, 5, 6, 7, 8, Squares, 1, 4, 9, 16, 25, 36, 49, 64, I Numbers, 9, 10, 11, 12, 13, 14, 15, 16, •^^^-^ Squares, 81, 100, 121, 144, 169, 196, 225, 256, Numbers, 17, 18, 19, 20, 21, 22, 23, 24, 25. Squares, 289, 324, 361, 400, 441, 484, 529, 576, 625. 2. Find the cubes of the integers from 1 to 10 inclusive, and commit them to memory.* C Numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. -^^*-) Cubes, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. *At the option of the teacher. EVOLUTION. 265 S. 922 _ ? 4. a)^ = ? 5, .32 — ? 6. (7|-)^ = ? '7. 3.082 zr: ? 8. .3712 _ ? 9. 43722 zzr? ^ 10. 5.82 = ? ai. 47.62—? 12. Q^)2 =. ? ^ws. S464. ^ns. If. Ans. .09. >4* Ans. 9.4864. ^ws. .137641. 13. (124f)2 = ? 14. 97^ = ? 15. 5.753 = ? 16. (3|)3z=:? 17. IF = ? 18. 1010 — ? 19. (4)12 = ? 20. .59 zn ? 21. Involve 1.3 to the 6tli power. 22. Raise 18f to the 5th power. 23. What is the difference between the square and the cube I CuU % of 24, ,Tr"'- ■ ^"- 1 ^ "^'"^ } '^^l Cu^^- - ' ^ -i "^ 24. What is the compound interest of $1.10 for 4 years, at 10 per cent ? J^\ J '':\ ^ j 25. How many paving stones 13 inches square will be required to pave 100 rods of a street 3 rods in width.'' 26. How many dice measuring ^ an inch each way may be made from a cubical foot iqf ivory, allowing -j^^ for waste in the manufacture? j 'i|i/ 1^ EVOLUTION. 983* Evolution *consists in finding the roots of numbers. 384* The root of a number is one of the equal factors which produce that number. The square root is one of the two equal factors ^ the cube root, one of the three equal factors ; the fourth root, one of the four equal factors, and so on. 385. a/ is the radical sign, and by itself signifies the square root, and with a figure above it, signifies the degree of the root indicated by the figure ; thus, ^^27 signifies the third root of 27. The root may also be indicated by a fractional exponent ; thus, 16^ (r«ad, 16 to the i power) = ^^ 16 =: 2. 5f66 EVOLUTION. SQUARE ROOT. 38G* Table, showing the places occupied by the square of any number of units, tens or hundreds. Koots. Squares. 1 squared =: j 9 " = 81 10 " = ioo 99 " = 980i 100 « = ioooo 999 " = 998001 38 T, By the above we perceive that the square root of any whole number expressed by one or two figures, must be units ; expressed by three or four figures, must be units and tens ; ex- pressed by five or six figures, must be units, tens, and hundreds. Plence, generally, if a number he separated into periods of two figures each, beginning with the units, the number of figures in the square root will be iridicated by the number of periods. Note I. — The left hand period may contain but one figure. Note II. — The principle above elucidated applies also to decimal frac- tions ; but every period in decimal fractions must contain two figures. 388. That the pupil may comprehend the method of ex- tracting the square root of a number, we will multiply 64 by itself, I. e., square it, and keep the separate products, instead of reducing them and adding as in ordinary multiplication. G4 X 64 = (GO + 4) X (60 + 4). (1.) 60 X 60 =: 602 — 3^00. (2.) {^4^60} =2 X (60 X 4)zn 480. (3.) 4X4= 42= 16. By the above it will appear that a square whose root is com- posed of tens and units, contains (1.) The square of the tens ; (2.) Twice the tens multiplied by the units ; and (3.) TRe square of the units. — 4096. SQUARE ROOT. 267 Operation. Trial divisor, 2 X 6(0) z= :12(0)\ 496 4 7 496 True divisor, .... 124 000 Or simply i 4096 ( 64 36 124 ) 496 496 389. We will now extract the square root of 4096. » i Separating the number HP into periods of two figures 4096 ( 64 each, we find that the root 6(0)2 ;— 36 ^j2 consist of two figures, and the square of the tens must be contained in the 40 (hundreds) ; the largest square contained in 40 (hundreds) is 36 (hun- dreds), the root of which is 6 (tens) ; this we write as the tens' figure of the root, and subtract its square 36 (hundreds) iGrom the 40 (hundreds), and to the remainder 4 (hundreds), bring down the next period, 96. This remainder (496) must contain two times the tens of the root multiplied by the units, pltcs the square of the units, or the jyroduct of two times the tens, plus the units, multiplied by the units. If it contained only two times the tens multiplied by the units, we should obtain the units' figure by dividing the remainder (496) by two times the tens. "We make 2X6 (tens) = 12 (tens) the trial divisor, which is contained in 49 (tens) 4 times. AVe write 4 as the units' figure of the root, and also at the right of the 12 (tens), and have 124 for the true divisor. This we multiply by 4, and thus complete the square, — obtaining at once, twice the product of the tens by the units, and the square of the units. If the root consists of more than two figures, having obtained tha ftrst two, we can consider them as tens in reference to the next figure, and proceed with them in all respects as above. Thus, suppose it be Operation. 4i38.94 ( 64.33-f 36 124 ) 538 496 1283 ) 4294 3849 12863 ) 44500 38589 5911 required to extract the square root of 4138.94: find the first two places as before ; bring down the next period, 94, and form a new trial divisor by doubling 64 (the root already found) ; find how many times this, consid- ered as tens, is contained in 429 tens, for the third 268 EVOLUTION. figure of the root. To obtain a fourth figure in the root; form another IDeriod by annexing two zeros, double 643, and so continue. From the above, we deduce the following Rule for extracting the Square Root of a Num- ber. Point off the given number into periods of two figures each, by placing a dot over the units' Jigure and every alternate figure to the left in whole numbers, and to the right in decimals. Find the greatest square number in the left hand period, and write its root as the first term in the answer. Subtract the square number from the left hand period, and to the remainder hring down the next period for a dividend. Take twice the root already found for a trial divisor ; rejecting the right hand figure of the dividend, divide it by the trial divisor ; place the result, as the second term in the root, also at the right of the trial divisor, mahing a true divisor ; multiply the true divisor thus obtained by the last term of the root, and subtract this product from the dividend ; to the remainder bring down the next period for a new dividend. Double the terms of the root already found for a new trial divisor, and proceed as before. Note I. — When a zero occurs m the root, annex a zero to the trial divisor; bring down another period, and proceed as before. Note II. — If a root figure proves too large, substitute a smaller, and repeat the work. Note III. — When a remainder occurs after all the periods are brought down, the root may be more nearly approximated by annexing periods of • zeros, and continuing the operation. Note IV. — The square root of a common fraction maybe obtained by ■ extracting the root of both terms when they are perfect squares ; when they are not, the fraction may first be reduced to a decimal. Note V. — Mixed numbers may be reduced to the decimal form, or to improper fractions when the denominator of the fractional part is a squar« number. SQUARE ROOT. 269 R Fig. 1, c 3600 sq. ft. s IS 60 ft. D Fig. 2. 390. The above rule may be illustrated by diagrams. Let A B C D (Fig. 1) represent a square court containing 4096 square feet, the length of whose side we wish to determine. Having found (Art. 389) that the greatest square of tens in 4096 is 3600, the root of which is 6 tens, we deduct 3600 from 4096, and have left 496 square feet, which are to be disposed on two sides of the square already found. The width of these additions we wish to ascertain. By extending the lines a and b, we shall divide the addition into three parts, M, N, and O ; M and N having for one side the tens of the root, and O being a square whose side is equal to the width of the side additions. If the 496 square feet equalled the feet in the side additions, M and N, the width of the ad- ditions would be determined by dividing 496 by twice the length of the square already found, 2 X 60. Using this as the trial divisor, we obtain 4 as the width, which is the units* term of the root ; but the entire length of the additions rig. 3. is two times the tens, plus the units, or 124 (Fig. 3), the product of which by 4, the units' term, is 496. 60 + 60 +4 There being no remainder, 4096 is found to be a square of which 64 is the root, and the length of the court is 64 feet. s 60 ft. B- M 391. Examples in Square Root. 1. "What is the square root of 841 ? Ans. 29. 2. What is the square root of 763876 ? Ans. 874. 13. What is the square root of 13616100 ? Ans. 3690. 4. What is the square root of 253009 ? Ans. 503. 5. What is the square root of 1012036 ? Ans. 1006. 0. AVhat is the square root of 447.3225 ? Ans. 21.15. .7. What is the square root of .005625 ? Ans. .075. 8. What is the square root of .169 ? Ans. .41109+. 9. What is the square root of ^^\ ? Ans. /^ 27Q EVOLUTION. 10. What is the square root of ^f ? Note.— 12— A Ans. ^. 11. What is the square root of lOy^^ ? Ans. S^, 12. What is the square root off? Ans. .86602-|-. 13. What is the square root of 8|? Ans. 2.8635-1-. 14 What is the square root of 9^9^ ? Ans. 3.02334-]-. 15. What is the square root of f of ||- ? Ans. ^^, Optional Examples. Note. — Extract the root in the following to five places. 16. ^2fxjHf = ? 17. V21025=:? 18. V9801^ = ? 19. V 502681=? 20. V22S=? 21. V14002564=:? 22. V-4X 25 =? 23. V2213.7025r=:? 24. V^^239 9025 z= ? 25. V2^ = ? 26. V4028049 = ? 27. V20j = ? 28. V 9569534976 = ? 29. Vl6X752rr:? SO. V gfHI|^ = ? 31. V^of f ofH = ? f 32. v-144:=:? i W- 33. V8 = ? 34. V81.10083136==? 35. V^=? 36. V746841.64r=:? 37. V T^gX( 6|)^ = ? 38. V769.987=:? 39. V^025 = ? 40. V/^ of .052 = ? 41. V.^ = ? 42. V(.25-^.06l)X(f)2=? 43. V 10011 = ? V S9S, Practical Examples. 1. There is a field of corn having an equal number of rows and hills in a row, which contains 1020100 hills in all ; what is the number of rows in the field? Ans. 1010 rows. 2. A body of troops, consisting of 2601 men, has an equal num- ber in rank and file ; how many are there in each ? Ans. 51 men. 3. A company of persons spent $3.24 ; each person spending as many cents as there were persons, how many cents did each spend ? Ans. 18 cents. SQUARE ROOT. 271 4. "What is the length of one side of a square farm containing S02 acres, 2 roods of land? Ans. 220 rods. 5. What is the length of a square park which contains 2 square miles ? A)is. 1-4142-|- miles. 6. There is a circular lot which contains 3 acres ; what is the length of a square lot whose ai'ea is the same ? Ans. 21.9089+ rods. 7. What is the size of a square lot whose area is thirty times that of the above? Ans. 120 rods. 8. What is the cost of fencing a square lot which contains 1 acre, at $5 per rod? Ans. $2 52.9S. 9. The side of a square is 8 ft. 6 in. ; what is the side of a square having 25 times the area ? - ' 10. A owned a lot of land 51 rods by 80 rods, and another 180 rods by 100 rods, which he bartered with B for a square lot con- taining 138 acres; how many rods less of fencing are there in the square lot than in the other two? Ans. 228 rods nearly. 11. I have two square lots of land, the larger of which con- tains 270 acres ; the ratio of the smaller to the larger is as 5 to 6 ; what is the length of one side of the smaller ? Ans. 189.73+ rods^ 12. On a roof there are laid 5000 slates, — the number in the length being twice the number in the breadth; wliat is the number each way ? ; Note. — It is evident that the slates are laid in two equal squares; hence the square root of ^ of 5000 (V^ of 5000) will equal the breadth. Ans. 50 sla*es in breadth ; 100 slates in length. 13. Suppose the above roof to have had 10000 slates, and the breadth to have been one third of the length, what would have been the number of slates in the length and breadth ? Ans. 173.205+ length; 57.735+ breadth. 14. What is the difference between the fencing of a 34-acre lot, whether it be a square or a rectangular lot, twice as long as it is wide? Ans. 17.89 rods. 15. My orchard contains 5400 trees ; the nupiber of trees in 272 EVOLUTION. width is to the number in lengfli, as 2 to 3 ; what is the number each way ? Note. — -| of the trees will be a square, whose square root will be the number of trees in the width of the orchard. 16. Suppose, in the above orchard, the outer rows of trees to stand upon the boundary line, and all to stand 30 feet apart, what is the area covered by the orchard? Ans. lOS^^f acres. 17^ There is a rectangular court paved with 1728 paving- stones 15 inches square ; the length of the court is to the width as 4 to 3 ; what is the number of stones each way ? 18. How many square feet in the superficial contents of the above court ? ^ ^ 19. What is the side of a square that will contain as many square feet as a rectangle w^hose sides are 150 and 70 feet ? 20. What is the mean proportional between 6 and 24? (Art. 373.) / APPLICATIOK OF SQUARE ROOT TO RIGHT-ANGLED TRI- ANGLES Definitions. 39S« An Angle is the opening between two lines that meet each other. 394:* A Right Angle is the angle formed by two lines that are perpendicular to each other, (Art. 191.) 39«>« A Triangle is a figure having three angles, and bounded by three straight lines. 396. A Right-angled Triangle is a triangle having one of its angles a right angle. 397. The Hypothenuse of a Right-angled Triangle is the side opposite the right angle. 398. The Base of a Right-angled Triangle is the side upon which it is supposed to stand. 399. The Perpendicular of a Right-angled Triangle is the side perpendicular to the base.- SQUARE ROOT 273 ~ VXA xy X; 5 •- 400. To FIND EITHER SiDE OF A RiGHT-ANGLED TRIAN- GLE, THE OTHER TWO SiDES BEING KNOWN. Suppose the figure A B C to be a right-angled triangle, whose sides are 3, 4 and 5 feet respectively. A square formed upon the hypothenuse, A C, will contain 25 square feet; one formed upon the base, B C, will contain 16 square feet, and one formed upon the perpendicu- lar, A B, will contain 9 square feet. Thus, it appears that the square upon the line A C is equal to the two squares upon A B and B C ; and generally, The square upon the hypothenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. Hence, Rule I. To find the hypothenuse, the base and perpendicular being giv^n : Square the base and perpendicular, and extract the square root of their sum. Rule II. To find the base or perpendicular, the hypothenuse and other side being given : Square the hypothenuse and the given side, and extract the square root of their difference. 4:01. Examples. 1. The base of a right-angled triangle being 30 feet, the per- pendicular 40 feet, what is the hypothenuse ? Ans. 50 ieet. 2. The hypothenuse of a right-angled triangle being 32.5 feet, the base 30 feet, what is the perpendicular? Ans. 12.5 feet. 3. What must be the height of the eaves of a house that can be reached by a ladder 30 feet long, the foot of the ladder stand- ing 18 feet from the underpinning of the house? Ans. 24 feet. '4. How far from the foot of a post 15 feet high can a horse feed that has a rope fastened around his neck and attached to the top of the post, the distance being 37 feet to the neck, and the horse feeding two feet beyond the end of the rope in a direct line with the rope ? Ans. 36 feet. 5. G. W. Bailey had a tree, which being partially broken off 18 274: EVOLUTION. 24 feet from the ground, the top struck the ground 10 feet from the foot of the tree, and on a level with it ; what was the height of the tree ? Ans. 50 feet. • 6. What must be the length of a ladder to reach to the top of a chimney 48 feet high, the foot of the ladder being 20 feet from the chimney ? Ans. 52 feet. 7. If the top of the ladder mentioned above be lowered 6 feet, how far will the foot stand from the chimney ? 8. Two vessels start at the same point, and sail, one due south 6 degrees, and the other due east 8 degrees ; how many miles apart are they, reckoning 69-^ miles to a degree ? 9. What is the width of a street from a point in which a ladder 32^ feet long will reach a window 26 feet high on one side, and one 24^ feet high on the other side ? Ans. 40.85-|-. 10. What is the width of a common, on which stands a flag- staff 195 feet high, from the top of which to one side of the common is 675 feet, and to the other 360 feet ? 11. How far from the foot of a flagstaff 24 feet high, must a ladder 23 feet long be placed that a person may ascend to within 5 feet of the top ? /A 12. My house is 40 feet wide, and the ridge-pole is 15 feet above the middle of the beam which connects the eaves ; what is the length of the rafters ? 13. Provincetown, Erie, and Elmira are in nearly the same latitude ; suppose Elmira to be 243 miles directly north of Wash- ington, Erie to be 305 miles north-westerly, and Provincetown 380 miles north-easterly, how far is Provincetown from Erie ? ) j' ^' 14. Four persons, Messrs. Ames, Barnes, Carnes, and Doane, are residing around Cincinnati, as follows : Ames, 20 miles north ; Barnes, 60 miles east ; Carnes, 27 miles south ; and Doane 36 miles west of the city ; what is the shortest distance one of these persons must travel to visit all the rest, and reach his own home ? <' / 15. What is the length of the diagonal, that is, the distance from one corner to the opposite corner, of a square lot which contains 16 square rods ? Ans. 5.6568-f- rods. SQUARE ROOT. 275 16t The diagonal of the floor of a square room is 110 feet; wliat is its area ? Ans. 6050 ft. 17. The diagonal of a square lot is 75 rods; what is one side ? 18, What is the diagonal of the floor of a room which is 15 fecit square ? . 19^ Suppose the above room to be 10 feet high; what is its diagonal, that is, its distiince from the lower corner to the oppo- site upper corner ? ^-f^oTE. — The diagonal of the floor (Ans. to Ex. 18) becomes the base of the triangle, of M'hich the diagonal of the room is the hypothenuse ; but the square of the diagonal of the floor is equal to the sura of the squares of the length and width of the room. Hence, to obtain the diagonal of the room, Square its three dimensions, and extract the square root of their sum, 20r What is the diagonal of a cubical room, each of whose dimensions is 20 feet ? Ans. 34.64-|- feet. 21*. What is the diagonal of a room 36 feet long, 24 feet wide, and 18 feet high? Z' /.^ 1/ 22? What is the diagonal of a cubical block whose edge is 2f inches? ^_^ 23.* In the centre of a square of land containing one acre is a mound 35 feet high ; at the top of this mound, which corresponds with the centre of the square, is a liberty-pole, 120 feet high; what is the distance from the top of the pole to the nearest point in the boundary of the lot? 24? What is the distance to the farthest point in the boundary J( / y^ 25? I have a lot of land 15 feet square, which I design to ar- range in five flower-beds as follows: a central square bed, to be bounded by lines connecting the middle points of the sides of the original square, and four equal triangular corner beds, whose sides extend 5^ feet from the right angle at the corner; how many feet of bordering will be required to surround all the beds? ^/25. 117.54 fl., nearly. 26? At the summit of a hill, which is 200 feet in height, stands a tower, 20 feet high ; from the top of the tower to the foot of the hill is 300 feet ; from the top of the tower to the opposita 276 EVOLUTION. side of a stream which flows at the foot of the hill is 400 feet ; what is the width of the stream ? S^^ For Dictation Exercises, see Key. )Ie, showing CUBE ROOT, the third po Eoots, 1 cubed — 9 « zrr 10 « nn 99 " — 100 " — 999 " — and hundreds. Cubes. i 729 iooo 970299 1000000 997002999 By the above it will be seen that the cube root of any whole number, composed of one, two, or three figures, must be units ; of four, five, or six figures, must be units and tens ; of seven, eight, or nine figures, must be units, tens and hundreds ; and hence, generally, that if we 'point a number off into periods of three figures each, heginning with the units, the number of figures in the cube root will be indicated by the number of the periods. Note I. — The left hand period may contain but one or two figures. Note II. — The principle above elucidated applies to decimal frac- tions ; but every period in decimal fractions must contain three figures. 4:03. Before extracting the cube root, let us involve 64 to the third power, and preserve the separate products. We have already seen (Art. 388) that the square or second power of 64 is 602 + 2 X ((30 X 4) + 42. By multiplying this square by 64 (60 r|- 4), we shall obtain the third power of 64. CUBE ROOT. 277 642 — 602 + 2 X ^QQ X 4) + 42 64 = 60 + 4 60^^2 X (60 X 60 X 4) + (60 X 4^) (60^ X 4) + 2 X (60 X 4^) + 4^ 643 — 60^ + 3 X (CO2 X 4) + 3 X (60 X 4^) + 4^ In this product we have four distinct parts, as follows : — (1), 603, the tens raised to the 3d power, z= 216000 (2), 3 X (602 X 4), 3 X the square of the tens X the units, =: 43200 (3), 3 X (60 X 42), 3 X the tens X the square of the units, = 2880 (4), 43, the units raised to the 3d power, =: 64 262144 Thus we see that a cube whose root is composed of tens and units, contains (1) the cube of the tens, (2) three times the square of the tens multiplied by the U7iits, (3) three times the tens multi' plied by the square of the units, and (4) the cube of the units. 4r04:, Observing, now, that the first of these parts is the cube of the tens, and that the units is a factor in each of the other parts, we will proceed to extract the cube root in the following III. Ex. What is the cube root of 264609.288 .? Operation. g| g . . . HPH 264609.288(64.2 Cube of tens, 603= 216 3 X sq. of tens (trial divisor), . . 3 X 6O2 = 10800 43509 * 3 X tens X units, 3 X 60 X 4 = 720 Square of units, 4' = 16 <■ True divisor, 11536 X 4= 46144 3 X sq. of tens (trial divisor), . 3 X 6402 = 1228800 2465288 3 X tens X units, 3 X 640 X 2 = 3840 Square of units, 22 = 4 True divisor, 1232644 X 2 = 2465288 Solution. Pointing the number oif into periods of three figures each, by placing a dot over the units and every third figure to the right and left, we find that the root will consist of three figures, and the cube of the tens must be contained in 264(000). The largest cube 278 EVOLUTION. contained in 264(000) is 216(000), the root of which is 6(0) ; this we write as the tens' term of the root, and subtract its cube, 216(000), from 264(000), and to the remainder, 48(000), bring down the next period, G09. We know that this remainder, 48609, contains three times the square of the tens (the term ah-eady found), multiplied by the units ; and though it contains other terms, since this is much the largest, we take 3 times 60^ (three times the square of the tens) for a trial divisor, and, dividing 48609 by it, obtain 4 for the units' figure. We multiply 3 X the tens (60) by the new term of the root (4), and place the product under the trial divisor, and under this, place the square of the units' figure ; and thus form our true divisor, consisting of the last three parts of a perfect cube (Art. 403), wanting the units Qs a factor in each. Multiplying their sum by 4, we have 46144. This we subtract, and to the remainder bring down the next period. Considering 64 as the tens in the root, we multiply its square (6402) by 3 for a new trial divisor, and, proceeding as before, obtain 64.2 foir the cube root of 264609.288. From the above we deduce the following 4©S. KULE FOR EXTRACTING THE CuBE EOOT. Point off the given number into periods of three figures eachy by placing a dot over the units, and every third figure to the left in whole nuni' bers, and to the right in decimals. Find the greatest cube in the left hand period, and write its root as the first term in the answer. Subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. Multiply the square of the root already found, considered as tens, by three for a trial divisor. Divide the dividend by the trial divisor, and place the result as the next term in the root. To the trial divisor add three times the former term in the root (considered as tens), multiplied by the last term, also the square of the last. Midtiply this sum by the last term, and subtract the product from the dividend. To the remainder bring down the next period for a new div- idend. Multiply the square of the terms of the root already found (consid' CUBE ROOT. 279 ere d as fens), hy three for a trial divisor, with which divide and proceed as before. Note I. — When a zero occurs in the root, annex two ciphers to the trial divisor, and, bringing down another period, proceed as before. Note II. — If a root figure proves too large, substitute a lower, and repeat the work. Note III. — AVhen a remainder occurs after all the periods are brought down, the root may be more nearly approximated by annexing periods of zeros, and continuing the operation. Note 1Y. — The cube root of a common fraction may be obtained by extracting the root of each of the terms when they are perfect cubes; when they are not, the fraction may be reduced to a decimal. Note V. — Mixed numbers may be reduced to decimal fractions, or to improper fractions when the denominator of the fractional part is a cube number. 4:06, The above rule may be illustrated by means oir* blocks. A cube number represents a cube, the edge of which is the cube root of the number. Let there be a cube of 262144 solid inches, whose edge we wish to determine. Having found by pointing and trial that the greatest cube of tens in 262144 is 216(000), the root of which is 6 tens, we will let 216000 inches be represented by the following diagram (Fig. 1), having for its edge 6 tens of inches, or 60 inches. Subtracting the cube, 216(000) in., from 262144 inches, there will remain 46144 inches, which may be disposed on three sides of the cube already found, so as to retain the cubical form. The square con- tents of the addition upon one side of this cube will be 60^ zzr 3600 inches, and upon three sides 10800 inches. Using this as a trial divisor, we find the thickness of the ad litions to be 4 inches. The additions are repiesented by Fig. 2. These additions being made, the solid will be represented by Fig. 3. Fig. 1. Fig. 2. ^^--^y -™5 pifiiiiiiiiiiiiiiiiiii B S III III "iinii 60 in. 4i n. 4 in. 4 in 280 EVOLUTION. Fig. 3. Fig. 4. Fig. 5. Fig. 6. To complete the cube, it also requires three oblong rectangular blocks, whose length is 60 inches, and whose end is 4 inches square (Fig. 4) ; also a cube, whose edge is 4 inches (Fig. 6). The side of one of the oblong blocks being 60 X 4, one side of the three will be 3 times 60 X 4 zzz 720 square inches, and one side of the small cube will be 42 z= 16 square inches. If, now, we multiply the sum of these surfaces, 10800 + 720 -f- 16, — j: 11536 (Fig. 7), by their thickness, 4, and increase the cube 216000 Fig. 7. 60 60 60 4 4 4 8 . 4 ~]4 3 X 602 -{- 3 X 60 X 4 + 4'-^ = 11536. Fig. 8. by the product, we form a perfect cube (Fig. 8), whose edge is 64 inches. And since thei-e is no remainder, 262144 is a perfect cube, of which 64 is the root. 1- C41u. 4:®7, Examples. 1. What is the cube root of 2744? 2. What is the cube root of 24389 ? . 3. What is the cube root of 704969? 4. What is the cube root of 12977875 ? 5. What is the cube root of 224755712 ? Ans. 14. Ans. 29. Ans. 89. Ans, 235. Ans. 608. CUBE ROOT. 281 6. What is the cube root of 122097755681 ? Ans. 4961. 7. What is the cube root of 729486108008? Ans. 9002. 8. What is the cube root of 19683000 ? A7is. 270. 9. What is the cube root of 195.112 ? Ans. 5.8. 10. What is the cube root of .000729 ? Ans. .09. 11. W^hat is the cube root of 329778750 ? Ans. 690.8+. 12. What is the cube root of .57? Ans. .8291+, 13. What is the cube root of 32^: ' Ans. 3.1854-. 14. What is the cube root of 4 ? Ans. 1.587-4-. 15. What is the cube root of tt¥¥ ? ^^s. j% 16. What is the cube root of A^ Ans. ^. 17. What is the cube root of A\5 ' (2¥, The area of a square or rectangle equals the of its length and its breadth or height. (Art. 173.) 4@6. The area of any parallelogram equals { — the product of its base and its height; for it can be proved to be equal to a rectangle of the same base and height. 427. The area of a triangle equals half^ the product of its base and height ; for every triangle equals one half of a parallelogram of the same base and height. product / AREAS. 285 When the three sides of a triangle are given, the area may be found hy subtracting each side separately from half the sum of the three sides, then multiplying the continued product of these remainders hy half the sum of the sides, and extracting the square root, 4:^8, The area of a trapezoid equals half of the sum of its parallel sides multiplied hy the distance between them ; M for it is equal to two triangles whose bases are the two parallel sides of the trapezoid, and whose altitude is the distance between them. N 429. The area of any polygon may be found by dividing it into triangles and obtaining the sum of their areas. Note. — The student should draw figures for each of the following problems. 430. Examples for Practice. 1. What is the area of a rectangle whose length is 20 feet and breadth 6j feet ? Ans. 130 sq. ft. 2. How many square feet of canvas in a picture 6 ft. 9 in. long and 4 ft. 2 in. broad ? J &" *^ 3. How many square yards in a garden 20 yards square ? 4. Required the area of a parallelogram whose base measures 3 ft. 4 in. and altitude 1 ft. 3 in. 5. Required the area of a parallelogram r -7 whose base measures 23 feet and the adjacent /_\ / side 13 feet, from the extremity of which a ^ perpendicular drawn to the base cuts from the base 5 feet. Ans, 276 sq. ft. 6. What is the area of a triangle, the length of the base being 20 feet, and the height 10 ft. 4 in. ? Ans, lO^ sq. ft. 7. What is the area of a right-angled triangle whose base and perpendicular are 20 and 18 feet respectively? / ^^ 8. What is the area of a right-angled triangle whose perpen- dicular and hypothenuse are 42 and 45 J- feet respectively ? \f Ans, 367^ sq. feet 266 MENSURATION. 9. Eequired the height and area of an equilateral triangle whose sides are 10 feet long. Ans. height 8.66+ ft.; area 43.3+ sq. ft. 10. Eequired the area of a triangle whose sides are 3, 8, and 10 feet long respectively. A7is. 9.921+ sq. ft. 11. How many square rods in a triangular lot of land whose jsides measure 14 rods, 32 rods, and 20 rods, respectively? / i '(^ 12. What is the area of a trapezoid whose parallel sides are 14 and 32 feet long, and the distance between them 16 feet? Ans. 368 sq. ft. 13. What is the area of a trapezoid whose parallel sides are twice those of the above, and the distance between them 5 ft. Tin.? . 14. How maiiy sq. ft. in the surface of a board which is 18 ft. long, 18 in. wide at one end, and 14 in. wide at the other? Ans. 24 sq. ft. 15. How many acres in a quadrangular field having 2 parallel sides measuring 10 ch. 5 1. and 16 ch. 8 1. respectively, and the distance between them being 15 ch. ? Ans. 19.5975 acres. 16. What is the area of a trapezium, the length of a diagonal being 50 feet, and of the perpendiculars from the opposite ver- tices to the diagonal 10 feet and 35 feet? Ans. 1125 sq.ft. 17. Find the area of the accom- panying polygon, the dimensions being as follows : A C, 5 ft. ; A D, 8 ft. ; A E, 10 ft. ; B M, 3 ft. ; C N, 4 ft.; DQ, 5 ft. 6 in.; P F, 4 ft. 6 in. J^i.h CIRCLES. 4:31* The area of a circle equals o??e half of the product of the circumference and radius, or one fourth of the product of the circumference and diameter ; for it may be considered as made up of triangles, whose bases compose the circumference ol tke circle, and whose vertices (Art. 190), are at the centre. CIRCLES. 287 43^* Geometricians have proved that the circumference of every circle is nearly 3.1416 times its diameter. Hence, When the Diameter is given, 433* The Circumference =: Diameter X 3.1416. 434. The Area — (Diametef X 3.1416) X ^^^^ — Diameter^ X .7854. When the Circumference is given, 435. The Diameter = «£^^™^ When the Area is given, 436. The Diameter =:^ 1 5I^^ \ .7851 437. Examples. I. Required the circumference of a circle whose diameter is 8 feet. Ans, 25.1328 ft. ^ 2. If a radius is 12 feet, what is the circumference? Ans. 75.398+ ft. 3. If the circumference is 100 feet, what is the length of the diameter? Ans. 31.8309+ ft, 4. What is the area of a circle whose diameter is 21 feet ? Ans. 346.3614 sq. ft. 5. What is the area of a circle whose diameter is 5 ft. 6 in. ? 6. What is the area of a circle whose radius is 2 ft. 1 in. ? Ans. 13 sq. ft. 91^ sq. in. 7. What is the area of a circle whose radius is 5 ft. 2 in. ? 8. What is the diameter of a circle whose area is 4 sq. rods ? Ans. 2.256+ rds. 9. What is the radius of a circle whose area is 19 sq. miles? Ans, 2.459+ miles. 10. What is the space occupied by a cart-wheel -whose spokes are 2 feet long, and the diameter of whose hub is 10 inches ? Ans. 18.3478+ sq. ft. II. How many square yards in a circular piece of cloth that will cover a haycock measuring from the ground over the top to the opposite side 10 feet ? Ans. 8.72| sq. yds. 288 MENSURATION. 12. How many sq. inches in the bottom of a square box that will contain a ring 20 inches in diameter ? Ans. 400 sq. in. 13. How many sq. inches in the bottom of a square box that will be contained in a circular box 20 inches in circumference ? P Ans. 20.263 sq. in. 14. How many rods squire is a plat of ground which contains as much as a circular plat that is 20 rods across ? Ans. 17.724-f rds. 15. How many planks 2 inches thick can be sawed from a log 10 feet in circumference, allowing 2 slabs, each at least 3 inches thick, to be cast aside? • Ans. 16 planks. Solids. ■P Cube. Parallelopiped. Pyramid. Cone. Cylinder. Frustum of a Cone. Sphere. DEFINITIONS, 289 » The Height of any of the solids here defined is the perpendicular -distance from the highest point to the base, (See lines A B in the preceding figures.) 440. The Slant Height of a Kegular Pyramid or of a cone is the shortest distance from the vertex to the perimeter (boundary) of the base. {See lines A m in the preceding figures.) 447. The Slant Height of a Frustum of a Pyramid or Cone is the shortest -distance between the perimeters of the two bases. (See lines o-p in the figures.) 448, A Olobe or Sphere is a «olid l^ounded by a curved surface, every part of which is equally distant from a point within called the centre, SOLIDITIES AND CONYEX SURFACES. 449« The Solidity of a Parallelopiped equals the product of its three dimensions^, (Art. 178.) 4^0» The Solidity of a Cube equals the cube of one of its edges. 19 290 MENSURATION. 4:51m T'le Solidity of a Prism or of a Cylinder equals the area of its base multiplied by its height ; for it is evident that a prism or cylinder 1 inch high must contain as many cubic inches as there are square inches in the base ; and if it is 2, 3, or any number of inches high, it must contain 2, 3, or that number of times as many solid inches. 4:53, The Convex Surface of an Upright Prism or Cyl- inder equals the perimeter of one of its bases multiplied by its height; for it is evident that, if the prism or cylinder is 1 inch high, its convex surface contains as many sq. inches as there are inches in the perimeter ; and if the prism or cylinder is any num- ber of times 1 inch in height, its convex surface must contain that number of times as many square inches. 4:«>3. The Solidity of a Pyramid or Cone equals the area of its base multiplied by -^ of its height ; for it can be proved that these solids are each ^ of a prism or cylinder of the same base and height. 4^4. The Convex Surface of a Pyramid or Cone equals the perimeter of its base multiplied by ^ of the slant height ; for the convex surface of each may be regarded as composed of triangles whose bases form the perimeter of the base of the solid, and whose height is the slant height of the solid. 4.^^* The Solidity of the Frustum of a Pyramid or Cono equals that of three pyramids or cones whose bases are the upper and lower bases of the frustum and a mean proportional (Art. 373) between the two, and whose height is the height of the frustum.. Hence, the solidity equals the sum of the iivo bases plus the square root of their product, multiplied by ^ of the height of ttie fruS' tum, 406. The Convex Surface of the Frustum of a Pyramid or Cone equals J- the sum of the perimeters of the two bases mul- tiplied by the slant height ; for the convex surface of each may be regarded as made up of trapezoids whose parallel sides form • the perimeters of the bases, and whose height is the slant height of the frustum. SOLIDITIES AND CONVEX SURFACES. 291 4l57» Geometricians have proved that the Convex Surface of a Sphere equals the circumference multiplied hy the diameter, or equals the area of four great circles * of the sphere, 458. The Solidity of a Sphere is equal to its surface mul- tiplied hy ^ of the radius, or -i of the diameter , for the sphere may be regarded as made up of pyramids whose bases comprise the surface of the sphere, and whose vertices are at the centre. From the preceding explanations, and by the use of the well estalDlished fact that the circumference of every circle is 3.1416 times the diameter, the following formulas for finding the solid contents and convex surfaces of cylinders, cones, frustums of cones, and spheres, are obtained. To save space, D will be used for diameter of lower base, D' for diameter of upper base, h. for height, and s. h. for slant height. 459. The SoUd Contentsof a Cylinder = D2 x .7854Xh. 460. The Solid Contents of a Cone = D^ X .7854 X-- 461. The Solid Contents of a Frustum of a Cone = (D2 X .7854 + D'2 X .7854 + D x D' X .7854) x\ = (D2 -I- D'2 -f D X D') X .7854 x|- 46^. The Convex Surface of a Cylinder = D X 3.1416 X h. 46S. The Convex Surface of a Cone = D X 3. 1416 X — 464. The Convex Surface of a Frustum of a Cone = (D X 3.1416 + D'X 3.1416) X '^•• 465. The Convex Surface of a Sphere = D X 3.1416 X D = D2 X 3.1416. .5236 466. The Solid Contents of a Sphere = B^ x ^.t4.t<^ X-j=D3x.5236. * A great circle of a sphere is a circle which divides the sphere into two equal parts. 292 MENSURATION. 467. Examples. -1. How many cubic feet does a block of granite contain, that is 12 feet long, 4 feet wide, and 1^ feet thick ? Ans. 72 cu. feet. 2. What number of cubic feet are there in a cube whose edge is 1 foot, 11 inches ? Ans. 7.041-|- cu. feet. 3. How many cubic feet in a prism whose base is a parallelo- gram 15 feet long and 4 feet wide, and whose height is 9 inches ? W 1^ / W^f ^'^^H=ljL) £/, -•,-:l.(l I- RELATIONS OF CIRCLES. 293 t. ^3. Required the capacity of a conical pit, measuring 8 feet across and 5 feet from the edge to the deepest part. Ans. 50.2656 cu. feet 14! How many quarts of water will a circular tin pan contain, that measures across the bottom 11^ inches, across the top 14 inches, the slant height being 3^ inches 7 Ans. 6.65-(- quarts. 15. How many cubic feet in a ball 5 feet in diameter ? Ans. 65.45 cu. feet. 16. How many square feet in the surface of the ball ? Ans. 78.54 sq. feet. 17. How many square inches of leather will cover a ball 4 inches in diameter ? 18. What proportion do the cubic contents of a cone bear to the contents of a cylinder which will just contain it ? Ans. \, 19. Wbat proportion do the cubical contents of a sphere bear to the contents of a cylinder which will just contain it? Ans. §. 20*. Suppose, when the moon is 238600 miles from the earth,t that its shadow just reaches the earth's surface, how many cubic miles in the shadow, allowing the diameter of the moon to be 21(5(X utiles, and that of the eai^th to be 8000 miles ? '^^"^^ •K- 4 I / ^ ^- ^A4.483,914,786,355.2 cu. miles. '^RELATIONS OF 61RCLES; STtMILAR TRtANGLES, AND' POLY- 4:G8. It will be apparent, by the annexed diagrams, that a figure 1 inch square will con- tain 1 square inch, one 2 inches square will contain 4 square inches, one 3 inches square will contain 9 square inches, and thus, generally, that the areas J GONS. 3 in. square. 2 in. square lin. D 1 sq. in. 4 sq. in. 9 sq. in. of squares are to each other as the squares of their edges. t The distance is measured from the centre of the earth to the centre of the moon. 294 MENSURATION. The same principle applies to circles, triangles, and all figures that are similar to each other;* hence, 4:69, I. The Areas of Similar Triangles and Polygons are to each other as the squares of their corresponding dimen- sions. III. Ex. A triangle whose base is 10 feet has an area of 15 feet ; what is the area of a similar triangle whose base is 12 feet? By Proportion, 102 ; 122— 15 ; 2I.6 square feet, Ans. 470, II. The Areas of Circles are to each other as the squares of their diameters, semi-diameters, and circumferences. III. Ex. If a pipe of 2 inches diameter will empty a cis- tern in 3 hours, what must be the diameter of a pipe to empty the same cistern in 1 J- hours ? By Proportion, 1 J : 3 =: 22 : 8, the square of the diameter of the re- quired pipe. v^8 zz= 2.828 + inches, A/is, 471* Examples. ~l. If the pot to a furnace which consumes 60 lbs. of coal a day IS 24 inches in diameter, what amount of coal will be consumed in the same time by a furnace whose pot is 15 inches, all other conditions being the same ? Ans. 23.437-|- lbs. 2. If a rope 3 inches in diameter weighs 20 lbs., what is the diameter of a rope of the same length which weighs 9 lbs. ? Ans. 2.012-fin. 3. If a pipe 4 inches in diameter fills a cistern in 20 minutes, 15 seconds, in what time will a pipe that is 2 J- inches in diameter fill the same cistern? Ans. 51.84 minutes. H 4. If it costs $10.50 to cover a roof whose length is 7 feet, what "will it cost to cover a similar roof whose length is 21 feet? Ans. $94.50. * Angular figures are similar when their angles are equal, and their corresponding sides proportional ; and, conversely, similar figures hare their corresponding sides proportional. SIMILAR SOLIDS. 295 5 The hypothenuse of a right-angled triangle is 40 feet ; what fliust be the hypothenuse of a similar triangle that it may contain twice the area? Ans. 5Q.5QS-{-. 6. If a circular lot of land which is 10 rods in diameter con- tains 78.5398 square rods, what number of rods will a lot contain which is o rods in diameter ? Ans. 19.63495. V7. The area of a triangle whose base is 24 feet is 120 feet; what is the area of a similar triangle whose base is 96 feet ? Ans. 1920 feet. •^' 8. The Winchester bushel is 18|- inches in diameter and 8 inches deep ; what must be the diameter of a circular measure 6 inches deep, that it may hold a bushel ? Ans. 21.36-|- inches. 9. I have a circular flower-garden, the circumference of which is boi^ered with 75 yards in length of sodding ; how many yards will be required to border a circular garden of f the area ? Ans. 61.237-f- yards. 10. Having a triangular board 7 J feet long, what distance from the base end shall I cut it to divide it into two equal parts ? J[ws. 2.197 — ft. SIMILAR SOLIDS. "Note. — Angular solids are similar when their angles are equal each to each, and arranged in the same way, and their corresponding edges proportionaL The following proposition may be easily proved by geom- etry : — 473o The Solidities of Cubes, Spheres, a7id all Similar Solids are to each other as the cubes of their corresponding dimensions. III. Ex., I. How man}'- globes of G inches diameter can be made from a globe of 48 inches diameter .^ By Proportion, 6^ : 48^=: 1 (globe) : 512 globes, Ans, III. Ex., II. If a conical stack of hay which contains f of a ton is 6 feet high, what is the height of a similar stack which contains 3f tons ? 296 MENSURATION. By Proportion, I : 3f = 63 : 1728, the ctibe of the height of the larger stack. ^1728= 12 ft., height of larger stack, Ans. Examples. 1. If an ounce ball is | inch in diameter, how many ounce balls can be made from a globe of lead G inches in diameter ? An&. SUj%% balls. 2. A pyramid which is feet in height contains 48 cubic feet ; what is the height of a similar pyramid that contains 100 cubic feet? Ans, UA9i-\- ieeL 3. If a cube of granite whose edge is 2 feet weighs 1336.32 pounds, what will be the weight of a cube whose edge is 4 feet, 9 inches? Ans. 17901.99. 4. If an egg of 2| inches in eircumference weigh 1 ounce, what would another of the same form and consistency weigli whose circumference is 6 inches ? 5. What must be the height of a cone t& contain I2i1 times as many solid inches as a similar cone 3 inches in height ? An», 15 inches. 6. If a bushel measure is 18^ inches in diameter and 8 inches deep> what must be the diameter and depth of a half-bushel measure similar in form ? Ans, Diam. 14.683+ in.; depth 6.349+ in. 7. If an elephant^s tusk 9^ feet long and 8 inches in diameter at base weigh 214 pounds, what would be the dimensions of a similar tusk weighing 75 pounds ? ( >,, / I 8. Estimating the mean diameter of the earth at 7912 miles, and that of the moon at 2160 miles, how many bodies of the size of the moon could be made fr^m the bulk of the earth ? 9. If the bulk of Saturn be 1000 times as great as that of the earth, what is the diameter of Saturn f / 10. At what distance from the top, parallel with the base, must a conical sugar-loaf 12 inches high be cut that it may he divided into two equal parts ? 11. Mr. Hoot has three stacks of hay of similar shape, the QUESTIONS FOR REVIEW. 297 * diameters of their bases being, respectively, 10, 12, and 14 feet; if the one whose diameter is 10 feet contains 2^ tons, what will each of the others contain ? ^/ ^ § C ^ !o l^ For Dictation Exercises, see key. ^ 4TS. Questions for Review. What is Involution. ? a power ? an exponent ? What is the first power ? second power ? third power ? fourth power ? What are the second and third powers generally called ? Rule for Involution ? How may a mixed number be raised to a given power ? Repeat the squares of the integers from 1 to 25 ;* the cubes of the integers from 1 to 10.* How does Evolution differ from Involution ? What is a root ? What is the square root of a number ? the cube root ? How is the square root indicated ? the cube root ? How other- wise may the root of a number be indicated ? If a power contain one or two figures only, of how many figures will its square root consist ? If a power contain three or four figures ? If five or six ? What three terms does every square number contain whose root consists of tens and units ? Give the rule for extracting the square root. Of how many figures may the left-hand period in whole numbers consist ? of how many must every period, except this, consist ? of how many, every period in decimals ? ''How do you proceed when a zero occurs in the root ? how when a root figure proves too large ? how when there is a remainder ? How do you extract the square root of a common fraction whose terms are square numbers ? whose terms are not squares ? How extract the root of a mixed number ? Explain the extraction of the square root by an example. Illustrate by diagrams. /"What is an angle ? a right angle ? a triangle ? a right-angled trian- gle ? its hypothenuse ? its perpendicular ? its base ? To what is the square on the hypothenuse of a right-angled triangle equal? Rule to find the hypothenuse ; to find base or perpendicular. If a cube number contain one, two, or three figures only, of how many figures will its cube root consist ? if it contain four, five, or si^ figures only? To what four terms is every cube number equal whose root consist! of tens and units ? * At the option of the teacher. 298 QUESTIONS FOR REVIEW- ♦ Give t*he rule for extracting the cube root Of how many figures may the left-hand period consist? of how many must every other period consist in whole numbers ? in decimals ? How do you proceed when a zero occurs in the root ? how when a root figure proves too large ? how when there is a remainder ? How do you extract the cube root of a common fraction when the terms are cubes'? how when the terms are not ? How extract the cube root of a mixed number ? Explain the extraction of the cube root by an example. Illustrate by blocks. What is Mensuration ? Name and describe the different kinds of triangles. Draw a right-angled triangle ; an obtuse-angled triangle ; an equilateral triangle ; an isosceles triangle ; a scalene triangle. Name and describe the different kinds of quadrilaterals. Draw a square ; a rectangle ; a rhombus ; a trapezoid ; a trapezium ; a circle ; a polygon of 5 sides with 2 diagonals. ' How do you find the area of a square ? a rectangle ? any parallelo- gram ? a triangle ? a trapezoid ? a trapezium ? any polygon ? How do you find the circumference of a circle when the diameter is given? when the radius is given? How do you find the area of a circle when the diameter is given? when the radius is given? How do you find the diameter when the circumference is given ? How find diameter when the area is given ? How do you find the radius when the area is given ? Define a cube ; parallelopiped ; prism ; cyhnder ; pyramid ; cone ; frustum of a pyramid or cone ; a sphere. Draw or mention some- thing in the form of each of these solids. What is the height of any solid ? the slant height of a pyramid or cone ? the slant height of a frustum of a pyramid or cone ? How do you obtain the solid contents of a cube ? a parallelopiped ? a prism or cylinder ? a pyramid or cone ? a frustum of a pyramid or cone ? How do you find the convex surface of each of these solids ? When the diameters and altitude are given, how do you find the solid contents of a cylinder ? of a cone ? of a frustum of a cone ? of a sphere ? How do you find the convex surface of a cylinder ? of a cone ? the frustum of a cone ? a sphere ? What proportion exists between the areas of squares ? of circles ? of all similar triangles and polygons ? When are angular figures similar ? What proportion exists between spheres ? between all similar solids? When are angular solids similar ? REVIEW. 299 474. General Review, No. 8. Pl. Supply the 2d term ia the proportion 3| : ? = 8 : 25 X G^/ 2 2. Wliat is the mean proportional between .8 and .72 ? ; J .^v^^ / 3. Divide $1900 between two men, in the proportion of 3 to 5.1 ^ 4. Divide $45 between three boys, so that one shall have as much as the other two. whose shares are as 2 to 7. y^J 5. If 15 gallons of oil cost 7 £, 10 s., what will 25 J- gallons cost ? .5 "^. .6. How many pounds can 5 horses draw, if 6 horses can draw as much as 10 oxen, and 2 oxen can draw 2400 pounds? •. 7. Smith and Lee formed a partnership. Smith put in $1000 for 6 months, and Lee $600 and his services for 8 months, his services being equal to $100 a month. They gained $1506 ; what was each one's share ? -— 8. What is the 5th power of 23 ? the cube of 96 ? 9. What is the largest number of men in a regiment of 1000 that can be arranged in a squarer; and how many men will re- main? How many men will there be on each side of the square ? 10. How many feet of fencing around a square farm containing 15 acres? 11. A ladder 27f feet long reaches a window 25f feet from the ground ; how far does the foot of the ladder stand from the house? ,4 :' / "■ ^12. Required the diameter of a circular grass plat which contains 314^^ square feet. :. 13. How many rods of fencing on both sides of a road which surrounds a circular park containing | of a square mile, the road being 3 rods wide ? 14. What must be the depth of a pail, that is 10 inches across, to contain 5 gallons (the sides being upright) ? 15. How many feet of canvas are required to construct a con- ical tent 14 feet across the bottom and 9^ feet from the highest point to the ground ? 16! How many gallons will a circular vat contain, which meas- ures across the top 8 ft., across the bottom 7 ft., the sides sloping uniformly'and measuring on the slope 6J- ft. in depth? ; ^^ For changes, see Key. / 'V ^ f \ / '^ 300 ALLIGATION. ALLIGATION. ^^5» Alligation, or Average, treats of the mixing of dif* ferent ingredients. 4:76. Alligation Medial is the process of determining the average or mean value of given quantities of different values. 4:77. Alligation Alternate is the process of determining what quantities of different values may be so combined that the mixture shall be of a given value. Note. — The word alligation means a tyinp together, and is applied to these processes because, in the solutions of many examples, the amounts or prices of articles are linked or tied together. Average is perhaps the better name to use, as it applies to all the examples. ALLIGATION MEDIAL, 478. III. Ex. Let it be required to mix 10 lbs. of sugar at 7 cents per lb. with 7 lbs. at 9 cents, and 8 lbs. at 11 cents ; what will be the value of the mixture ? Operation. The price of 10 lbs. at 7 cents per 10 X 7 = 70 lb.'= 70 cents; of 7 lbs. at 9 cents 7 X 9 ziz 63 =63 cents ; of 8 lbs. at 11 cents = 8 X 11 = 88 88 cents. Adding, we find the value 25 )221 of the mixture to be 221 cents, and "^ cents, Ans, *^^ ^"°^^^* °^ P«""^^' ^« ^' -^' ^^ ^°^ 25 lbs. are worth 221 cents, 1 lb. is worth -^ of 221 cents =: 8 ^2^1^ cents, Ans. Hence we deduce th^ fol- lowing Rule. To find the mean value of given quantities of different values : Divide the sum of the values of the several quantities hy the sum of the quantities. Examples. 1. If 10 lbs. of raisins worth 10 cents per lb. be mixed with 4 lbs. worth 15 cents per lb., what is the value of the mixture per pound? Ans. 11^ cents. ALLIGATION MEDIAL. 301 ^ 2. There are in a certain school, 10 pupils 14 years old; 9 pu- pils 12 years old; 5, 11 years; 8, 9 years, and 17, 10 years old; what is their average age ? "^■^3. A family spent, during the year, as follows : in January $89.75, in February $70.16, in March 185.32, in April $90.21, in May $87.00, in June $66.14, in July $69.42, in August $72.68, in September $80.65, in October $90.45, in November $98.54, in December, $109.63 ; what was their average expense per month? 4-.-'5t V ■ "^ 4. In Philadelphia, during the year 1861, rain or snow fell as follows : in January on 13 days, in February on 9 days, in March on 9, in April on 9, in May on 13, in June on 15, in July on 14, in August on 12, in September on 6, in October on 10, in No- vember on 11, in December on 4 ; what was the a\|erage number of days per month when rain or snow fell ? I v j-^ 5. In Massachusetts, during the year 1850, the value of home manufactures was $205,333. During the year 1860, it was $245,886. What M-as the average rate of increase per year during the 10 years ? q (J b^-/:>\J P 6. A flour merchant sold 50 bbls. flour at $7.50 per bbl., 60 bbls. at $9.00 per bbl., 25 bbls. at $8.50, 40 at $8.75, and 100, at $9.50; what did his sales average per barrel.'* V ': \ '•-,'{ )) 7. A baker made wedding-cake of the following ingredients : 5 lbs. flour worth 5 cents per lb., 5 lbs. sugar at 11 cents per lb., 5 lbs. of butter at 22 cents per lb., 6 lbs. raisins at 17 cents per lb., 12 lbs. currants at 20 cents, per lb., 2 lbs. citron at 50 cents per lb., 50 eggs, l^lbs. to the dozen, 18 cents per dozen, ^ pint wine at 374 cents per pint, 3 oz. cinnamon at 56 cents per lb., 3 oz. nutmegs at $1.00 per lb., 1^ oz. mace at $1.00 per lb. Al- lowing $2.00 for labor and fuel, -^ lb. for the weight of the wine, and 1 oz. in every lb. for loss of weight in baking, what was the cost of the cak« per lb. ? Ans. $.24^^* | ^. 302 ALLIGATION. V. ALLIGATION ALTERNATE. 479. III. Ex. A merchant has teas of the following val- ues per lb., 42,- 68, 75, and 84 cents, with which he wishes to make a mixture worth 70 cents per pound. How many pounds of each kind shall he take ? 70 42 -{- 28- 68 -|- 2-| Operation. lib. 5 lbs. S-" 2 lbs. 14-^ 2 lbs. A71S. 1 lb. at 42, 5 lbs. at 68, 2 lbs. at 75, and 2 lbs. at 84 cents. We first compare the various prices of the tea with the price ot the mixture. If that which is worth 42 cents is sold at 70 cents, there is a gain of 28 cents on one pound, which we indicate by writing -(-28 opposite 42. In the same way we find there is a gain of 2 cents per lb. on the 68-cent tea, a loss of 5 cents per lb. on the 75-cent tea, and a loss of 14 cents per lb. on the 84- cent tea. We indicate the gain and losses by their proper signs, and proceed to take, two by two, such kinds of tea and of such quantities that the gains shall balance the losses. Comparing the first with the fourth, we find that the gain on 1 lb. of the first equals the loss on 2 lbs. of the fourth. We also find' that the gain on 5 lbs. of the second equals the loss on 2 lbs. of the third. We therefore take 1 lb. of the first, 5 of the second, 2 of the third, and 2 of the fourth ; or, we may take any quantities of the first and fourth that are in the ratio of 1 to 2, and of the second and third that are in the ratio of 5 to 2. Instead of comparing the first with the fourth, and the second with the third, we may compare the first and third and the second and fourth together, thus : — 70 42 + 28--, 5 lbs. 68 -f- 2-f -1 7 lbs. 75 _ 5-1 I 28 lbs. 84 — 14 — ^ 1 lb. Other comparisons might be made, and thus an indefinite number of answers be obtained. But it is best to compare those gains and losses together that have the greatest common factors; for in such com- parisons, whatever factors are common can be disregarded, and the remaining factors of each gain or loss will show the required quantity ALLIGATION ALTERNATE. 303 of the other artfcle. From the above operations we derive the fol- lowing Rule. To find whtxt quantities of different values shall be taken to make a mixture of a given value : Write the different values in a column with the medium value at the left. Compare each given value with the value of the mixture. Write what it requires to equal that of the mixture in a column at the right with the sign -\- prefixed, or what it exceeds that of the mixture with the sign — . Take such quantities of each ingredient that the gains and losses shall he equal. 4:80. Proof. Examples in Alligation Alternate may be proved by finding the mean value of the several ingredients as given in the answer, and comparing it with the given mean value. 481. The following simple method of solving this class of examples is sometimes given, which is preferable whenever it does not give fractional portions of the given quantities. III. Ex. Let it be required to mix sugar at 8, 9, 11, 13, and 15 cents per lb., that the mixture may be worth 10 cents per lb. 8-f2 Gain on lib. =2 * This deficiency of 6 in the 10 ^ 11 — 1 Loss on 1 lb. r= 1 gain may be made up either by taking G lbs. more of the 9-cent sugar, or 3 more of the 8-cent, or 2 more of each of the 8 and Sum of losses = 9 9-cent sugars. Deficiency of gain = 6 First answer, 7 lbs. at 9 cents, and 1 lb. of each of the others. Second ansiver, 4 lbs. at 8 cents, and 1 lb. of each of the others. Third answer, 3 lbs. at 8 and 9 cents, and 1 lb. of each of the others. 48^. Examples. 1. How shall corn at 50 cents a bushel, be mixed with grain at 80 cents a bushel, that the mixture may be worth 75 cents. per bushel.? Ans. 1 bu. at $.50 to 5 bu. at $.80. 9 + 1 <( (( lib. = 1 11 — 1 Loss on lib. — 1 13 — 3 (( (( lib. — 3 15 — 5 a (( lib. = 5 Sum f gains — 3 §04 ALLIGATION. , 2. How shall oil at 80, 95, and |1.50 per gallon, be propor- tioned that the mixture may be worth $1.00 per gallon? 3. How shall tea at 62, 75, 68, 90, and 98 cents, be propor- tioned that the mixture may be worth 80 cents per lb. ? 4. A grocer makes a mixture of syrup, worth 62 cents per gall., from syrups worth 45, 60, 75, and 80 cents per gall. ; how many gallons of each may he use ? 5. A grocer has cider at 28 and 30 cents per gall., which he wishes to mix with vinegar at 27 cents per gall., and water, so that the mixture may be worth 25 cents per gall. ; what propor- tions may he use ? Ans. 1 gal. of each of the other ingredients to f gal. water, etc. 483. When one of the quantities is limited, Jind the entire gain or loss on that quantity, and take such quantities of the other ingredients that their gains and losses shall balance each other and the gain or loss on the limited quantity. When more than one quantity is limited, Jind the resulting loss or gain from taking the limited quantities, and balance as before. III. Ex. How much tea at 60, 75, and 87 cents per lb., may be mixed with 30 lbs. of tea at 95 cents per lb., that the mixture .shall be worth 85 cents per lb. ? " Operation. 60 + 25 X 12 = -f 300 75-l-lO-j 1 87— 2^ 5 1^95 — 10 X 30 = — 300 Ans. 12 lbs. at 60 cents, 1 at 75 cents, and 5 at 87 cents. Examples. 6. Plow many lbs. split peas at 5 cents per lb., must be put with 40 lbs. coffee at 21 cents per lb., that the mixture shall be worth 14 cents per lb.? Ans. 31 1 lbs. 7. A goldsmith has gold 16 carats fine, which he wishes to mix with 4 oz. gold 17 carats fine, 5 oz. 20 carats fine, 2 oz. 22 carats fine, and 3 oz. 24 carats fine, that the mixture may be 18 carat! fine ; how many oz. of it shall he use ? -t n) 85 ARITHMETICAL PROGRESSION. 305 Note, —The term carat is a word used in indicating the proportion of pmre gold in any given quantity of the metal ; thus, if the metal be pure gold, it is said to be 24 carats fine; if two thirds gold, 16 carats fine ; if 17 parts gold and 7 parts alloy, 17 carats fine, etc. 8. How much wool, of equal quantities, at 35 and 40 cents per lb., must be mixed with 100 lbs. at 60 cents per lb., that the mix- ture may be worth 45 cents per lb. ? 4:84:« When the entire quantity is limited, ^w^ the proportion of the ingredients as before, and then divide the given quantity among the ingredients in the proportion found. Examples. 9. J. Blake has an order from New York for 1000 bushels of wheat, at $1.25 per bushel. How shall he mix his wheat, which he values at $1.20, $1.22, and $1.30, to fill the order? Ans, 100 bu. at $1.20, 500 bu. at $1.22, 400 bu. at $1.30. '10. J. Smith wishes to purchase a farm of 200 acres, at $100 '^ an acre. How much woodland at $125 per acre, mowing up- ^ land at $90 per acre, pasture land at $70 per ^cre, and tillage ground at $128 per acre, may he purchase?) ^ w ->/' * ' L\(J ^h 11. How many lbs. of cotton at 60, 73, and 98 cents per lb., must be mixed with 750 lbs. at 90 cents, that the mixture may contain 2000 lbs. at 80 cents per lb. ?. A/O Vr /i A ^Omy^'^. Note. — First balance the loss on the 750 lbs. with gaiit on one of the-^^ > other ingredients taken; then proceed to make a mixture of the other ^i . ingredients equal to the entire quantity given, minus the quantities bal, N V ajiced. ' I ARITHMETICAL PROGRESSION. 485. Arithmetical iProgression is progression by equal differences. 486. An Arithmetical Series is a succession of numbers which increase or decrease by a common difference. If the numbers increase from the first term, the series i« an Increasing Series: e. g., 2, 4, 6, 8, 10, 12, &c. 20 306 ARITHMETICAL PROGRESSION. If the numbers decrease from the first term, the series i? a Decreasing Series; e. g., 13, 11, 9, 7, 5, &c. ^rST, In every series, five tilings are to be considered; viz., the First Term, the Last Term, the Namher of Terms, the Common Difference, and the Sum of the Terms ; any three of which being given, the other two may be found. Tiiis gives rise to twenty distinct cases, a few of the more important of which will be here presented. Note I. — For the remaining cases, also for full discussions of Geo- metrical Progression and Annuities, the student is referred to works on Algebra. Note IT. — Increasing series only will be considered in this book, as rules that apply to increasing series apply to decreasing series also, pro- vided that, wherever the common difference is introduced, it is used with the contrary sign. 488. To FIND ANY Term in a Series, when the First Term, Common Difference, and Number of Terms are given. Let 5 = first term, 2 = common difference, and 6z=the number of terms. The series will be constructed as follows : — (1.) Ist term. 2d term. 3d term. 4th term. 5th term. 6th term. 5. 5 + 2. 5 + 2X2. 5 + 3X2. 5+4X2. 5 + 5X2. We find that the second term equals the first term, plus the common . difference ; the third term equals the first term, plus two times the com- mon difference ; the fourth term equals the first term, plus three times the common difference, &c. ; and that the last or sixth term equals the first term, plus five times the common difference. Hence, I. To find any term of the series : Add the first term to the 'product of the common difference multiplied hy the number of terms which precede it. II. To find the last term: Add the first term to the product of the common difference multiplied hy the number of terms less one. ARITHMETICAL PROGRESSION. 307 Examples. 1. In an increasing series the first term is 4, and the common difference is 8 ; what is the seventh term ? Atis. 52. 2. The first term is 7, the common difference ^, and the.4iura- ber of terms 20 ; what is the last term ? / 5 T 3. If 5 lbs. of power is imparted to a fly-wheel at eachl-evolu- tion, what is its power at the end of the tenth revolution from a ' state of rest, provided its average loss of power from friction an^ ,, other causes is 1 lb. during each revolution ? Ans. 40 lbs. "" r 4. If a stone, in falling to the earth, descends IGyi^- feet during the first second, 3 X 16^^ feet during the next, 5 X 16iV feet during the third, and so on ; how far will it fall during the elev* enth second ? j > ) f- 5. What is the amotmt of $200 at pimple interest for 8 years, at 6 per cent. ? ^ f b ^ " ' •' '-' ■ ^^"^ ■ H ^ r • ' . Note. — The amount will be the ninth term of the series, of which the first term is $200. 4:89. To FIND THE Common Difference in a Series, ALSO THE Number of Terms. If, in series (1.) we subtract the first term from the last, we have r8o maining 5X2, that is, the common difference multiplied by the number of terms less one. Hence, I. To find the common difference : Divide the difference be- tween the first and last term by the number of terms less one. II. To find the number of terms : Divide the difference between the first and last term by the common difference, and add one to the quotient. Examples. 6. The first term of a series is 7, the last term 19. and the number of terms 13 ; what is the common difference ? Ans. 1. 7. The first term is 30, the last term is 3, and the number of terms 10 ; what is the common difference ? J 8. The first term is 8, the last term 23, and the common dif- fe>^nce 1^ ; required the number of terms, f | 308 ARITHMETICAL PROGRESSSION. 9 A boy, in picking up stones 2 feet apart, and carrying them, one at a time, to a deposit 2 feet from the first, found that to carry the last one, he had walked 60 feet ; how many stones did he carry in all? Ans. 15 stones. 490. To FIND THE Sum of the Series. Let 2, 4, 6, 8, 10, 12, 14, 16, be a series, of which we wish to find the sum. We write under it the same series in an inverted order, an^ add the terms as follows : — 2 4 6 8 10 12 14 16 16 14 12 10 8 6 4 2 18 18 18 18 18 18 18 18 We then have the sum of both series z= 8 X 18, or the sum of one series in: s.J:yiK But 8 equals the number of terms, and 18 the sum of the extremes. Hence, To find the sum of a series : Multiply one half the sum of the extremes hy the number of terms. Examples. 10. The first term of a series is 4, the last 40, and the number of terms 11 ; what is the sum of the series ? Ans* 242. 11. What is the sum of the odd numbers from 1 to 99 inclu- sive? /i . ] ^(r6' 12. What is the sum of the multiples of 3 from 6 to 45 in- elusive ? 13. How many notes must a person sing in ascending two octaves, if he goes back to the first note each time he strikes rt new one, and sounds all the intermediate notes each time he ascends? Ans. 120 notes. 14. Two of Dio Lewis's pupils tried their skill in running for pegs. Each set up 5 pegs 6 feet apart, and commenced running 6 feet from the first peg. How far did each run to place the pegs at his starting-point ? / ^ '" 15. How far would the first boy of a row of 21 scholars travel, Jn gathering writing-books from the row, if the scholars were 2^ feet apart, and he brought one book at a time to his own desk ? . GEOMETRICAL PROGRESSION. 309 GEOMETRICAL PROGRESSION. 4:91. Geometrical Progression is progression by equal multipliers. 4:9^. A Geometrical Series is a succession of numbers which increase or decrease by a common multiplier. Thus, 2, 4, 8, 16, 32, 64, is an increasing geometrical series, in which the multiplier is 2. 2) Ij h h h tV' ^s ^ decreasing geometrical series, in which the multiplier is ^. 4.93. The common multiplier is called the Katio. 494. In every geometrical progression, five things are to be considered ; viz., the First Term, the Last Term, the Number of Terms, the Common Eatio, and the Sum of the Terms ; any three of which being given, the other two may be found. 4:9«>. To FIND THE Last Term of a Seriks, the First Term, the Ratio, and Number of Terms being given. Let 3 be the first term, 2 the ratio, and 5 the number of terms. The series will then become, (1.) 1st term. 2d term. 3d term. 4th term. 5th term. 3, 3X2, 3X2^ 3 X 2^ 3X2^ in which the second terra equals the first term multiplied by the ratio, the third term equals the first term multiplied by the second power of the ratio, the fourth term equals the first term multiplied by the third power of the ratio, and the fifth term equals the first term multiplied by the fourth power of the ratio. Hence, I. To find any term of the series : Multiply the first term hy the ratio raised to a power equal to the number of terms which precede the required term. II. To find the last term of the series : Multiply the first term hy the ratio raised to a poioer equal to the number of terms less SXO GEOMETRICAL PROGKESSION. Examples. /^ 1. What is the seventh term of the series 2, 6, 18, 54, &c.?/ L^f ■ — T^lT^' ^^^^ ^^ *^^^ fifteenth terra of the series 5, 2^, 1 J, |, -j^^^, &c. ? r-^ ^' 3. What is the amount of $500 for 7 years, at 6 per cent., compound interest ? Note. — 1.06 is the ratio, and the amount the eighth term of the series. — 4. Naturalists have found that the ratio of increase of some kinds of animalcute (microscopic animals) is often four in a single day. At that rate, what would be the increase of one animal- cula and its descendants in ten days? Ans. 1,048,576. 4:96* To FIND THE Ratio, the First Term, the Last Term, and Number of Terms being given. In series (1), if the last term, 3 X 2^, be divided by the first term, 3, the quotient will be 2\ or the fourth power of the ratio, the fourth root of which will equal the ratio. Hence, To find the ratio : Divide the last term hy the first term, and extract that root of the quotient whose index equals the number of terms less one. Examples. 5. The first term of a series is 2, the last term 128, the num- ber of terms 3 ; what is the ratio ? Ans. 8. 6. The first term is 4, the last term J, and the number of terms 4 ; what is the ratio ? Ans. ^. 7. The extremes are 5 and 625, and the numbt^r of terms 4 ; what is the ratio? Oii, , * ^;49T, To FIND the Sum of a Series. Let 3, 9, 27, 81, 243, be a series,' of which we wish to find the sum. OPERxiTIOX. 3 times the first series = 9 27 81 243 729 ; subtracting the first series =i 3 9 27 81 243, we have twice the first series = — 3 729, or 729 ■— 3. .-. the first series z::^ i^^Tj^^^. ANNUITIES. 314 By multiplying each term of the series by the ratio, 3, we have a second series, \vhose sum is 3 times that of the first series, from which we subtract the first series ; the remainder equals twice the sum of the first series, wliich we find by dividing by 2. Hence, To find the sum of the series : Subtract the first term from the product of the last term multiplied by the ratio ; divide the remain- der by the ratio less one. Note. — If the series is descending, the last term multiplied by the ratio should be taken from the first term, and the remainder be divided by one less the ratio. Examples. 8. What is the sum of the series 3, 12, 48, 192, 768, 3072? Ans. 4095. 9. The first term is 5, the last term 3125, and the number of terms 5 ; what is the sum of the series ? Ans. 3905. 10. What is the sum of 7 terms of the series 4, 8, 16, 32, &c. ? 11. If 1 of the air in a receiver be taken from it by an air- pump at the first stroke of the piston, and ^ of the remainder at the second stroke, and so on, what will be the amount taken from the receiver by 8 strokes ? Ans. flf. ANNUITIES. 408. Annuities are periodical payments of fixed sums of money, in consideration of money paid or services rendered. 409. When an annuity is made for a definite number of years, it is called a certain annuity ; when it is made forever, a perpetuity ; when it depends upon the life of one or more per- sons, a life annuity ; when it does not commence till a given time has elapsed, it is said to be in reversion. ^00. When annuities are granted by government, they are called Pensions. 0O1. The Amount of an annuity is the sum of ^11 the pay ments, plus their interest, from the time they become due. 312 ANNUITIES. «>0d. The Present Worth of an annuity is such a sum of money as, put at interest, will exactly pay the annuity. ^03. Annuities are said to be in Arrears, or Foreborne, when they remain unpaid after they become due. 504:, Annuities are generally computed at compound inter- est. ANNUITIES AT SIMPLE INTEREST. «>0^. III. Ex. What is the amount of an annuity of $200 a year, at 6 per cent, simple interest, 5 years in arrears ? The payment due at the end of the fifth year is $200 ,• that which was due at the end of the fourth year amounts, at the end of the fifth year, to $200 plus the interest on the same for 1 year; that which was due at the end of the third year, to $200 plus its interest for 2 years ; that due at the end of the second year, to $200 plus its interest for 3 years ; that due at the end of the first year, to $200 plus its interest for 4 years. Hence, the sums due at the end of the fifth year would form an arithmetical series, 200, 200 + 12, 200 + 24, 200 + 36, 200 + 48, of which the first term is $200, the last term the amount of $200 for the number of years less 1, and the number of terms the number of years. Hence the sum may be found by Art. 490 ; and, generally, To find the amount of an annuity at simple interest : Find the sum of an arithmetical series, of which the first term is the last payment, the last term the amount of the frst payment, and the number of terms the number of payments. Examples. 1. What is the amount of an annuity of $300 for 6 years, at 6 per cent., simple interest ? Ans, $2070. 2. What is the amount of an annuity of |600 for 7 years, at 7 per cent, simple interest ? 3. A gentleman's salary of |?12O0 a year, payable quarterly, remained unpaid for 4 years ; what was then his due ? ANNUITIES AT COMPOUND INTEREST. ^06« III. Ex. What is the amount of an annuity of |36 for 4 years, at 6 per cent., compound interest ? ANNUITIES AT COMPOUND INTEREST. 313 We will first find the amount of an annuity of $1 for the same time. The last payment, due at the end of 4 years, will be $1. The sum due on the third payment, at the end of the fourth year, will be the amount of $1 for 1 year; that due on the second payment will be the amount of $1 for 2 years; that due on the first payment will be the amount of $1 for 3 yeai's. Hence the four sums due will form the geometrical series, 1, 1.06, 1.1236, 1.19101, or 1, 1.06, 1 X ^1.06)2 1 X (1.06)3, of which the first term is the last payment, the last term the amount of the first payment, and the number of terms the number of payments. Finding the sum of this series (Art. 297), and multiplying by 36, we obtain the required amount. Hence, the Rule. To find the ^amount of an annuity at compound inter- est : Find the anwunt of an annuity of $1 for the given time by geometrical progression (Art. 497), and multiply the sum thus C9. Bre(M and lioring traded in liideS for one ydar'.' ^Brbck put^ in $2000 at first ; at the end of 3 months he withdrew $700, and at the end of 7 months put in $1000. Loring put in $1200 at first, and $500 more in 4 months. At the end of 6 months he withdrew $200. The gain for the year was $2355.75, of which Loring received $1000 for conducting the business. What was the share of each ? ,> ,,^, - ^/ />^<^? f, I ' ' 72. Required the cost of boards, at $20 per thousand feet, to make a box 7 it. 10 in. long, 3 ft. 8 in. wide, and 2 it. 6 in. high ; boards to be 1 in. thick. Ans. $2.20§. 73. A wine merchant used the following receipt for port wine: 35 gallons prepared cider, worth $1.00 per gallon ; 5 gallons red wine, at $2.00 per gallon ; 5 gallons port wine, at $5.00 per gal- lon ; 3 gallons spirits, at $1.00 a gallon ; 3 pounds sugar, at 16 cents per pound ; 2 ounces tincture of kino, at 6 cents an ounce ; MISCELLANEOUS||;XAMPLES. 323 i and Leunce tartaric acid, at 13 cents. Suppose the sugar, kino, and acid do not add to the bulk of the mixture, and that the mer- chant sells it at $4.50 per gallon, what per cent, does he gain ?„ ,-j )jiV 74. If a pipe 2^ inches in diameter, will fill a cistern in two '' ' hours, in what time wilra'prp^'S'iiffcheii in cl^ame'ter fill the same? ,— 75. AVhat is the length of the edge of the largest cube that can be sawed from a globe 9 inches in diame^'? ly ^ j 'I -- 76. The ridge-pole of a house is 46 ft. from the ground, th* eaves 38 ft., the rafters on each side of the roof being 18 ft. long; what is the width of the house? 77. Required the edge of a cubical box that will contain 12a times as much as a box measuring 1 foot each way. ' 78. The pyramid of Cheops, in Egypt, is said to contaiiv 82111000 cubic feet of masonry, and to ha'Te been 480 feet high. Allowing 7000000 cubic feet, which are required to perfect its pyramidal form and to fill its cliambers, whfft is the length of ont* ' side of its base, which is a square ? Ans. 746.2 ft.-j-. 79. How many yards of cloth | yd. wide will be required i& cover the sides and top of a cubical box containing 6751.269 cubic inches ? 80. What will be the cost of digging a ditch outside a square garden containing 12.75 rods, the ditch to be 7 ft. wide and 5 fit, deep, at 1 cent per cubic foot ? ; ■ 81. How much would the earth taken out of the ditch raise the ^ surface of the garden ? , [ -f- 82. How many gallons in a cylindrical jat, the radius of whose base is 1 foot, and whose altitude is 4 ft. ? / '-Ul , JJ 83. "What will be the thickness of a square stick of timber which contains 4.542098 tons (50 cu. ft. =-z 1 ton), the stick being 100 ft. long? 84. Supposing a cubic foot of snow to weigh 31 lbs., what will be the pressure of a body of snow 9 inches deep upon a flat roof 100 fl. by 25 fl. ? ^ n I I / 85. Required the number of square feet in the surface of a ditch surrounding a circular garden which is 25 yards across, the ditch being 24- ft. across. L, l ':i 'k 324 miscellan;^ug examples. 8G. How many gallons v/ill the above ditch contain, it being o ft. deep ? ^^-^ i h' ' ■. -• /' 87. The circular outlet to a cistern being 4 inches in diameter, ■vvhat must be the width of a rectangular receiving-pipe, whose depth is 2 inches, that its capacity may be the same as the dis- charging-pipe ? . "■■^88. What will a pine log weigh whose length is 18 ft., meas- \iring 3 ft. across the larger end, and 2^ ft. across the smaller, pine being .6 as heavy as water, which weighs 62 J lbs. to a cubic foot ? (See Art. 455.) ^ 89. How many cubic yards in a cellar whose side walls meas« lire, on the outside, 70 ft., and whose end walls measure 48 ft., the cellar being 10 ft. deep, and the walls 3 ft. thick 7 // ^ 4, '90. Suppose there is a globe of ice in the region of the Alps weighing 243474 lbs.; there being 930 oz. to a solid foot, what is its diameter ? ' Ans. 20 ft. 91. Suppose ^ of the above globe of ice to melt away each yeaiv,>v.hat will be the length of its diameter each succeeding year ?**/ k r 92. An engineer planted a battery near the bank of a river to shell a fort upon the opposite side. To ascertain the distance of the fort, he noted the direction of the fort from the mortar ; then, placing himself at a point eight rods higher up the river, he caused a line to be drawn from a point six feet distant from himself, ia range with the mortar, to be extended parallel with the line first noted till it ranged between himself and the fort. This line he found to be 480 feet. What was the distance of the fort from the mortar ? (Page 294, Note.) Ans. 2 miles. 93. Wishing to know the height of a flagstaff which was 60 feet distant, I held my cane perpendicularly so that its lower end was 2 ft. 4 in. in a horizontal line from my eye, and found the range of the top of the staff was 35 inches from the bottom of the cane. Required the height of the flagstaff, allowing my eye to have been 5 ft. from the ground, which was a horizontal plane, Ans. 8) ft. high. APPENDIX. SOME OF THE PROPERTIES OF 9. «511. Any number may he separated into two parts, one of which is divisihle hy 9, and the other of which is equal to the sum of its digits. Illustration. Let 5864 be the number considered. 5000 = 5 X (999 I H- 60 = 6 X l-l- 4 = 1) — 5 X 999 + 5 i)z=8X 99 — 8 (9-j-l)=:6X 9 + 6 4 5864=: (5X 999 + 8 X 99 + 6 X 9) + (5 + 8 + 6^ 4). .-.5864 is separated into two parts, the first (5 X 999 + 8 X 99 + 6 X 9) being divisible by 9, and the second (5 + 8 + 6 + 4) being the sum of its digits. The same can be shown of any number. The following principles are derived directly from Art. 511 : — 31^. Any number is divisible hy 9, if the sum of its digits is divisible by 9. •ilS* If any number is divided hy 9, the remainder is equal to the remainder when the sum of its digits is divided hy 9. ^i4:. Pkoof of Multiplication by casting out the 9's. (See Art. 50.) Proof. 3 + 2 + 6 = 0+2 4 +2=_6 12. 1 + 2 r= 3, remainder. 1 + 3 + 6+2 = 12 = + 3, remainder^ These remainders being equal, the work is probably correct. (325) Multiplication. 326 42 652 1304 13692 326 APPENDIX. Demonstration of Proof. 326 -f- 9 leaves an excess of 2. We separate the mul- 42 -^ 9 leaves an excess of 6. tiplicand and multiplier Q2g 094. I 2 each into two parts, the 4.2 :::z 36-4-6 ^^^^ part being divisible 324X6 + 2X6 ^^ ^' ^''^ ^^^ '^""''^ 324 X364-2X36 P^^^ hemg the excess of — ; 9's. Multiplying- these 324 X 36 + 324 X 6 + 2 X 36 + 2 X 6 „„^,^,.^ ^^ =^p^^^. ted, we obtain four terms for a product, the first three of which are divisible by 9, and the last is the product of the two excesses. The entire product divided by 9 must, therefore, leave the same remainder as the product of the excesses in the multipHcand and multiplie/ divided by 9. 515, Proof of Division by casting out the 9's. (See Art. 62.) Division. • Proof. 75 ) 3929 (52 7_)_5 = 12=:0+3 375 5 + 2= 7^ "l79 21 2 + 1 = 3, remainder. 150 3929—29 = 3000, 3, remainder. 29 These remainders being equal, the work is prob- ably correct. Demonstration of Proof. The dividend, minus the remainder, equals the product of the divi- sor and integral part of the quotient ; therefore, it v e divide the divi- dend, minus the remainder, by 9, the remainder thus obtained must be the same as that which results from dividing the product of the excess of 9's in the divisor and the integral part of the quotient by 9. CONTRACTIONS IN MULTIPLICATION AND DIVISION. Arithmetical operations may sometimes be shortened mate- rially by the use of contractions in Multiplication and Division. A few have been suggested in Articles 52, 53, and 64. Some additional contractions are here given, which pupils are cautioned against using until they are so familiar with the common methods as to make no mistakes. APPENDIX. 327 516. To Multiply by 9, 99, 999, &c. 9 being one less than 10, 99 one less than 100, and 999 one less than 1000, &c., To multiply by any number whose terms are all 9's : Annex as many zeros to the ^hultiplicand as there are 9's ifi the multiplier^ and from that product subtract the multiplicand; thus, 27 X 99 = 2700—27 = 2673. Examples. 1. 36 X 99=? 2. 264 X 999 = ? 3. 58 X 9999=? 4. 36841 X 9999? 5. 7 X 9999999 = ? 6. 245 X 999999 = ? 7. 241 X 998 = ? (241 X 998 = 241 X 1000 — 241 X 2.) Ans. 240518. 8. 356 X 9995 = ? | 9. 54932 X 999997 = ? 517, To Multiply by a Composite Number, i. e., by a Number that is itself the Product of two or more Numbers. Separate the multiplier into convenient factors, multiply the midtiplicand hy one of the factors, and that product by another factor, and so on, till all the factors are employed ; the last prod* uct is the true answer ; thus, 41 X 25 := 41 X 5 X 5. Examples. 1. Multiply 368 by 72 ; by 36. 2. Multiply 4079 by 81; by 48. 3. Multiply 2145 by 108 ; by 144. 4. Multiply 50411 by 55 ; by 150. 518. To Multiply by Aliquot Parts of 10, 100, 1000, &c. Multiply hy 10, 100, 1000, <^c., as the case may require, and then find the required part ; thus, to multiply by 5, multiply by 10, and divide the product by 2 ; to multiply by 25, multiply by 828 APPENDIX. 100, and divide by 4 ; by 125, multiply by 1000, and divide by 8; by 33^, multiply by 100, and divide by 3 ; by IGf, multiply by 100, and divide by G ; by 12^, multiply by 100, and divide by 8. Examples. 1. 8743008 X 5 = what? 2. 8003478 X 25 = what ? 3. 786342 X 12^ = what ? 4. 875402 X 3^ = what? 5. 1090806 X 16|= what ? 6. 543297 X 125 = what ? A ^19* To Divide by a Composite Number. III. Ex. Divide 390 by 15. Operation. To divide by a composite number : Sepa- 3 ) 390 rate the divisor into convenient factors, divide 5 ) 130 hy one factory and the quotient thus obtained by 26, Ans. aii-other factor, and so on till all the factors are employed. The last quotient is the answer required. Examples. 1. Divide 243873 by 32. 2. Divide 39726 by 18. 3. 8954^121 = what? 4. 49176 -^- 72 z= what? «S^0. To find the True Remainder. III. Ex. Divide 83248 by 84. Operation. In this example we have remain- 3 ) S3242 Rem. ders after the several divisions. 4)27747 1. 1 The first remainder is of the same 7 ) 6936 3. 3 X 3 = 9 denomination as the first dividend, 990 6 6X12 — 72 °^ units. The second remainder rr> • -I "TTZT ^ of the Same denomination as the 1 rue remainder, 82 i v • i j o- o .u ' second dividend, or 3 s. 3 threes p:=. 9 units. The third remainder is of the same denomination as the third dividend or 12's. 6 twelves := 72 units. The entire remainder equals the sum of these several remainders, or 1 -[-9-f-72i=82. Hence, To find the true remainder : Commence with the remainder resulting from the second division, and multiply each partial re- mainder by all the preceding divisors except the one which gave that remainder, and add the sum of the products to the remain- h'r resulting from the first division. (if'r APPENDIX. 329 Examples. 1. 86543-^-117 = what? 2. 234567 -^ 324 = what ? 3. 359762-^187 = what? 4. 32947-^-132 = what? 5. 927638 -f- 2800 = what? 6. 7362851 -^ 693 = what? •521. To Divide by Aliquot Parts op 10, 100, 1000, &c. To divide by 5, divide by 10, and multiply the quotient by 2 ; to divide by 25, divide by 100, and multiply by 4; by 125, divide by 1000, and multiply by 8 ; by 33^, divide by 100, and multiply by 3 ; by 16§, divide by 100, and multijjly by 6 ; by 166§, divide by 1000, and multiply by 6, &c. Examples, 1. 9876 -^ 25 = ? 2. 34543-^-125=? 3. 87096 -M6f = ? 4. 432872 ^12i = ? 5. 687904-^250=? 6. 110748 -M662 = ? MODES OF ESTIMATING THE TIME BETWEEN TWO DATES. 5^fi, The mode adopted in this book for estimating the time between two dates, when interest is computed in months and days, is in common use among business men, and is the one most consistent with the ordinary method of computing interest at 30 days to the month. ^23* Another mode of estimating the time between two dates, is to find the number of entike calendar months, and count Jhe remaining days. «]»S4. A third mode of finding the time between two dates, consists in counting the exact days from one date to the other. ILLUSTEATION. By the first of these modes, if a note for months, which matures* Feb. 10th, 1865, were discounted at a bank the 11th of December previous, the time would be estimated at 1 day less than two months j viz., 1 month and 29 days. By the second mode, it would be estimated at 1 month and 30 days, i. e.y 2 months. ♦ A note is said to mature when it becomes due. 330 APPENDIX. By the third mode, it would' be estfuiated" at 61 days, or 2 months and 1 day. In the above, and in all similar cases, the second of these methods will give more time than the first, and the third will give more time than the first or second. If, however, the above note matured March 10, and were discounted Jan. 11, whilst the time by the first mode would be 1 day less than 2 months, — viz., 1 month, 29 days, — by the second mode it would be 1 month, 27 days, and by the third mode, 58 days, cr 1 month, 28 days. In this case, the third mode gives less time than the first, and the second, less time than the third or first. These differences arise from the difference in the length of the calendar months ; and, since a majority of the months have 31 days, the second and third modes will, on the whole, give more time than the first, and the third more time than the first or second. To avoid these irregularities, it is customary to make notes, running for short times, payable in 30, 60, or 90 days, instead of 1, 2, or 3 months. Note 1. — By custom, at the banks, a note which is given for months matures on the day corresponding with its date : if the month in which it matures has no corresponding day, it matures on the last day of the month. Thus, four notes dated severally the 28th, 29th, 30th, and 31st of Dec, 1864, and given for two months, all mature Feb. 28, with grace, March 3 ; while one dated Feb. 28 would mature April 28 and May 1. Note 2. — A note falling due on the Sabbath, or on a legal holiday, must be paid on the business day next preceding. Thus, when a holiday occurs on Monday, notes maturing that day must be paid on the pre- vious Saturday. To Compute Interest for Exact Days. S^5» The interest for days is calculated, in some of the tlnited States, in Great Britain, and by the United States gov- ernment, at 365 days to the year; 1 day's interest being consid- ered al^ of a year's interest. By the ordinary method it is calculated at 360 days to the year; this gives a year's interest /or gea of a year, which is ^f 5, or j^^ too much. Hence, To obtain the true interest for days : Subtract from the interest found by the ordinary method, -^ of itself , APPENDIX. 831 A TABLE Showing the Number of Days FROM ANY . TO THE SAME DAY OF DAY OF Jan. Feb. Mar. Apr. May. June. July. Aug. Sept. Oct. Nov. Deo. January, 365 31 59 90 120 151 181 212 243 273 304 334 February, 334 365 28 59 89 120 150 181 212 242 273 303 March, 306 337 365 31 61 92 122 153 184 214 245 275 April, 275 306 334 365 30 61 91 122 153 183 214 244 May, 245 276 304 335 365 31 61 92 123 153 184 214 June, 214 245 273 304 334 365 30 61 92 122 153 183 July, 184 215 243 274 304 335 365 31 62 92 123 153 August, 153 184 212 243 273 304 334 365 31 61 92 122 September, 122 153 181 212 242 273 303 334 365 30 61 91 October, 92 123 151 182 212 243 273 304 335 365 31 61 November, 61 92 120 151 181 212 242 273 304 334 365 30 December, 31 62 90 121 151 182 212 243 274 304 335 365 Note. — In leap years, if the last day of February is included in the time, a day must be added to the number obtained from the table. MEASUREMENT OF LUMBER. ^^ »30. The contents of boards, and of hewn and round timber, whether they are of uniform dimensions throughout or taper reg- ularly, may be found by rules explained in Mensuration. The following additional directions for finding their contents may be serviceable : — Board Measure. ^ST, If a board be 1 inch or less in thickness, its contents are found hy multiplying its length hy its mean breadth. The mean breadth of a board tapering regularly is half the sum of the breadth of its two ends. If it is irregular in shape, the average of a number of measurements at equal distances must be used as the mean breadth. If the board is more than 1 inch thick, the 832 APPENDIX. square contents in feet must he multiplied hy the thickness in inches: Thus, a board whose length is 7 feet, mean breadth 2 feet, and thickness ^ inch, contains 14 feet, board measure; but a board whose length and breadth are the same as given above, and whose thickness is 2 inches, contains 28 feet, board measure. Examples. 1. Required the contents of a board 10 ft. long and 1 in. thick, which is 2 ft. wide at one end, and 1 ft. 6 in. wide at the other. 2. Required the contents of a board 14 ft. long, J in. thick, and measuring at the ends 2 ft. 3 in. and 1 ft. 9 in. respectively. 3. Required the contents of a board 16 ft. long and 1^ in. thick, Tvhose mean breadth is 2 ft. 5 in. MEASUREMENT OF TIMBER. 528. The contents of timber are often estimated in board measure, and are found bi/ multiplying the product of the length and mean breadth expressed in feet, hy the mean depth expressed in inches. If estimated in cubic measure, the product of the length, mean hreadth, and mean depth, expressed in the same dimensions, must he found. Examples. 1. Required the contents of a piece of joist 20 ft. long, whose mean breadth is 4 inches, and mean depth 3 inches, board meas- ure. '^ i ^^2. Give the contents in cubic measure. / ; ' =^ / ' ^ ^/3. How many feet, board measure, in a stick of timber 15 ft. long, the breadth at the ends being 6 in. and 4 in., and the depth at the ends 4 in. and 2^ in. respectively ? 5S^, The contents of round timber may be obtained hy as- certaining its mean diameter, andfro?n that, as a hasis, estimating its contents as if it were a cylinder, or hy squaring I of the mean girt, and multiplying the square hy twice the length. 1. How many cubic feet in a stick of round timber, whose mean gh-t is 8 ft., and whose length is 12 ft. .'' APPENDIX. r»33 MISCELLANEOUS. ^SOa Shingling, and other plain work, as flooring and par- titioning, are generally estimated by the square of 100 feet. 1000 shingles are allowed for a square. «$ Paving is measured by the square foot, or square yard. ^S5* Bricklaying is generally estimated by the thousand bricks ; sometimes it is estimated by the square yard, square rod, or square (of 100 ft.), allowing 1^ bricks, or 12 in. in thickness. A great variety of methods for measuring prevail. Some workmen make no allowance for doors and windows, others make allowance of half the space occupied by doors and windows, and others still estimate the exact amount of material and labor em- ployed. Measurements are taken on the outside of walls, no allowance being made for corners. In estimating the number of bricks used, an allowance of one tenth of the solid contents is made for mortar. GAUGING. •>36* Gauging is the process of finding the capacity of casks, or other vessels, in gallons or bushels. cSvlT. To find the capacity of a cask or barrel : Add the squares of the head diameter, of the bung diameter, and of twice the middle diameter. Multiply the sum thus obtained by .0005667 ^me5 the length,and theresult willbethe contents ingaUons. Note. — The middle diameter is the diameter of the section midway between the bung and the head. The dimensions used in the above rule should be expressed in inches. 1. Find the capacity, in gallons, of a cask whose length is 40 inches, the head diameter 25 inches, the bung diameter 82 inches, and the middle diameter 29 inches. Ans. 113.634 gal.-f- 334 APPENDIX. New Hampsiiike Rule fou Annual Interest with Partial Payments. t5S8. AVhen notes are given upon "annual interest" (Art. 277), and partial payments are made during the year, instead of following the United States rule (Art. 274), which makes a new principal at the time of each payment, — 1. Compute annual interest upon the princijoal to the end of the first year in which any payments are made ; also compute interest upon the payment or j^o^ynients from the time they are made to the end of the year. 2. Apply the amount of such payment or payments, frst to cancel any interest that 7nay have accrued upon the yearly interests, then to cancel the yearly interests themselves, and then toioards the payment of the principal. 3. Proceed in the same icay toith succeeding jyayments. 4. If, at the time of any payment, no interest is clue except what is accruing during the year, and the payment or payments are less than the interest due at the end of the year, deduct such payment or pay- ments at the end of the year, loithout interest added. Note. No interest should be computed beyond the time of settlement. ^00. Examples. 1. A note for $2500, dated Oct. 1, 1860, was given on demand, with interest payable annually at 6 %. r June 1, 1862, $500.00. Endorsements, \ March 17, 1863, 87.94. (jDec. 1, 1865, 1000.00. What was due April 1, 1867 ? Operation. Principal, $2500.00 Oct. 1, 1860. Yearlv 5 i"te^est, 150.00 to Oct. 1, 1861. ' ^ ) interest, 150.00 to Oct. 1, 1862. Sunple interest on { ^ ^ 5 from Oct. 1, 1861, $150.00, 5 ^l!^ \ to Oct. 1, 1862. Amount, 2809.00 Oct. 1, 1862. Payment, 500.00 June 1, 1862. Interest, 10.00 to Oct. 1, 1862. Amount of payment, 510.00 Principal, 2299.00 Oct. 1, 1862. APPENDIX. 335 PrjNCirAL bro't forw'd $2299.00 Interest, 137.94 to Oct. 1, 1863. Payment, 87.94 March 17, 1863. Balance yearly int., 50.00 due Oct. 1, 1863. Simple interest, 9.00 to Oct. 1, 1866. C interest. 137.94 to Oct. 1, 1864. Yearly j interest, 137.94 to Oct. 1, 1865. ( interest. 137.94 to Oct. 1, 1866. $137.94 at simple int., 24.83 497.65 for 2 yrs. -\- 1 yr. == 3 yrs. Amomit, 2796.65 Payment, 1000.00 Dec. 1, 1865. Interest, 50.00 to Oct. 1, 1866. Amount of payment, ~ 1050.00 Principal, 1746.65 Oct. 1, 1866. Interest, 52.40 to April 1, 1867. Balance due, 1799.05 (( « Afis. $1799.05. 2. A note for $4200, given May 27, 1862, was payable on demand, with interest at 6 % annually; what was due on the note May 27, 1867, the following payments having been made: Pavments $ ^"-- ^' ^^^^' ^^^•'^2. raymenis, j j^^^ ^7, 1867, 3000.00. Ans. $2500.48. 3. A note given for $5000, on demand, at 6 % annual interest, was dated January 1, 1860, and endorsed as follows: July 1, 1860, $50. July 1, 1865, $500. " 1, 1861, 500. " 1, 1866, 2500. What remained due after the last payment was made ? Ans. $3703.62. 040. To Compute Interest at 7y^^ %. Multiply the principal hy twice the numher of days, and point off four places if the principal contains only dollars^ six places if it contains cents. Examples. - 1. What is the simple interest of $100 for 12 days, at 7y=V % ? Ans. $.24. .-^2. What is the simple interest of $300 for 9 days, at 7tV % ? ! Ans. $.54. X 336 THE METKIC SYSTEM. THE METRIC SYSTEM OF WEIGHTS AND MEASURES. Note. — The Metric System of weights and measures was first adopted in France in 1795. A length supposed to be one ten millionth of a quadrant, or one forty millionth of a circumference of the earth meas- ured over the poles, was taken as a provisional measure for the base of the system ; this length was called a Meter. In order to ascertain more accurately the length of a quadrant, new measurements of the earth were subsequently instituted under the direc- tion of eminent mathematicians, who measured the arc of a meridian between the parallels of Dunkirk and Barcelona. From their measure- ments, the length of the meter now in use was determined. This length was adopted as the base of the system, in 1799. The use of the metric system was not, however, legally enforced, to the exclusion of any other system, until January 1, 1840. In Spain, Portugal, and Belgium, this system is also used exclusively, while in many other countries it is adopted wholly or in part. Among tliese are Holland, Italy, Greece, Austria, Switzerland, and Poland, in Europe, and Mexico, Chili, Venezuela, Brazil, Ecuador, Guatemala, San Salvador, and the Argentine Republic, on this continent. Movements are also being made to adopt it in England, Germany, Sweden, and Norway. Its use in the United States was legalized by an act of Congress, passed in July, 1866. Notwithstanding so much has been done to make the meter exactly one ten millionth of a quadrant, it is now thought to be too short by a small fraction, which is, however, less than one eight thousandth of an inch. The length of the meter is nearly 39.37079 English inches, or 39.3685 United States inches ; but for ordinary purposes, may be con- sidered 39.37 inches. 454:1 • The Metric System is so called from the Meter, which is the base of all the weights and measures which it employs. . The Meter is the primary unit of length, and equals about 39.37 inches, or nearly 3 ft. 3| in. Upon the Meter are based the following primary units : the Square Meter, the unit of measure for small surfaces ; the Are the unit of land measure ; the Cubic Meter, or Stere, the uni of volume ; the Liter, the unit of capacity ; and the Gram, ih& unit of weight. From these primary units the higher and lower orders of units are derived decimally. i MEASURES OF LENGTH. 337 54:^0 The names of the higher orders of units are formed by prefixing to the name of tlie primary unit the following, from the Greek numerals : — Deka (10), Hecto (100), Kilo (1000), Myria (10000). The names of the lower orders of units are formed by pre* fixing to the name of the primary unit the following, from the Latin numerals : — Deci (lOth), Centi (100th), MilU (1000th). Consequently, the word dekameter signifies ten meters ; deka- litevy ten liters ; hectometer, one hundred meters ; hectogram^ one hundred grams ; kilometer, one thousand meters ; myria- meter, ten thousand meters, etc. So, also, the word decimeter signifies the tenth part of a meter ; centigram, the hundredth part of a gram ; milliliter^ the thousandth part of a liter, etc. MEASURES OF LENGTH. Note. — In this table, and in those which follow, the name of the pri- mary unit is designated by capitals, and the names of other important units by italics. 543. Table. 10 mil'limeters Q"^'"^) =. 1 centimeter, marked ("'"). 10 cen'timeters = 1 decimeter, " ^decim\ 10 dec'imeters = 1 meter, " (™). 10 METERS =1 dekameter, " ^dekam\ 10 dek'ameters =: 1 hectometer, " C"^)» 10 hec'tometers = 1 kilometer, " C^)' 10 kil'ometers = 1 myriameter, " /myriam\ Exercises. 1. How many meters equal 1 dekameter ?/■' 1 hectometer? 1 kilometer h q 1 myriameter ? 2. How many dekameters equal 1 hectometer? 1 kilometer? 3. 1 meter equals how many decimeters? how many centi- meters ? how many millimeters ? 22 338 THE METRIC SYSTEM. K IN' o — o 04:4:. The outer diagram in the margin represents a measure 4 inches in length ; the inner diagram represents a measure 1 decimeter or 10 centimeters in length. These diagrams will enable the pupil to compare the units of length of the metric system with those in common use. Note 1. — The new five- cent piece (of 1866) is 2 centuneters hi diameter. # Note 2. — 25 millimeters, or 2-^ centimeters, nearly equal 1 inch. Note 3. — 5 meters are nearly equal to 1 rod. Note 4. — 1 kilometer is a little less than |^ of a mile. Note 5. — 1 myriaraeter is a little more than 6^ miles. Although the meter is generally con- sidered the unit of length, yet in esti- mating great distances, as the length of a road, of a river, the distance between two cities, etc., the kilometer is regarded as the unit ; thus, the length of the Ohio River is 1528 kilometers, the distance from Troy to New York is 267 kilo- meters. 54:5. The manner of writing the different orders of units of length is illustrated by the following Table. 1 i i s P. 1 i 1 CO o p 0) a B 6 1 1 1 1 1. 1 1 1 MEASURES OF LENGTH. 339 In writing numbers by the metric system, the decimal point is usually placed at the right of the figure denoting the primary unit ; thus, the number 5 meters, 9 decimeters, is written 5.9™. If in writing a number any intermediate orders of units are wanting, their places should be supplied by zeros ; thus, 1 deka- meter, 2 millimeters, is written 10.002™. Exercises. Write the following in figures : — -1. Three (J^kameters, four meters. Ans, 34™. ^2. Seven hectometers, three meters. Ans, 703™. ^3. Three hectometers, one dekameter, five meters. i / ^\, -^4. Three kilometers, two hectometers, seven meters. 3 ^ - 5. Nine myriameters, five hectometers. / ^ v ^ --6. Two meters, four decimeters. Ans. 2.4™. ~ 7. Two meters, two centimeters, four millimeters. ""8. Five dekameters, two decimeters, eight centimeters. 54LO, When numbers are expressed by figures, the part of the expression at the left of the decimal point is usually read in the denomination of the primary unit ; the part at the right of the decimal point may be read either as a decimal part of the unit, or in the denomination indicated by the place of the last figure. Thus, in reading the expression 34.62™, we may say either 34 and 62 hundredths meters, or 34 meters and 62 centimeters. Exercises. Read the following, giving the name of each order of units : — 1. 23™. 2. 25.1™. 3. 321.05™. 4. 7137.008™. 5. Read the above in the denomination of the units as indi- cated by the abbreviation'™. 6. Read the same, giving each decimal part the denomination indicated by the place of the last figure. Since the metric system is a decimal system, a number expressed in units of one order may be reduced to units of another order by multiplying or dividing by ten, or some power 340 THE METRIC SYSTEM. of tea. If the number is written in figures, it is only neces- sary to remove the decimal point to the right of the figure indicating the required order, and give the expression its proper abbreviation ; thus, 59.36'" may be reduced to 5.936^*=^^", .59GG''''", .05936'^'", .005936"^^"*'", 593. 6*^^""^, 593G."", 59360.""". 547. Examples. 1. Express 5.24 meters as centimeters. Ans. 524*''". 2. Express 37.2 meters as kilometers. Ans. .0372^"*. 3. Express 12 hectometers as meters. Ans. 1200™. 4. Express 25 millimeters as meters. Ans, .025"^. 5. In 518 meters how many decimeters? how many centi- meters ? how many millimeters ? 6. In 3687 metres how many dekameters? how many hec- tometers ? how many kilometers ? 7. Express in meters the following, and add them: 4075 centimeters, 27 dekameters, .075 kilometers. Ans. 385.75'". 8. Express in kilometers the following, and add them : 2400 meters, 500 dekameters, .79 myriameters. Ans. 15.3^™. 9. If 7.08 kilometers are taken from 42 kilometers, how many meters will remain ? Ans. 34,920". 10. The distance round a certain park is 2.58 kilometers ; how many meters will a man go who rides around it six times? Ans. 15,480"". 11. A schoolboy walked one third around the above park in 1 2 minutes ; how many meters did he walk in 1 minute ? Ans. 71.66"^+. 12. The latitude of Chicago is 42° N. ; how many kilometers is it from Chicago to the equator? Ans. 46 6 6. 6 e'""'-}-. V. MEASURES OF SURFACE. 5J4:8. In the measurement of small surfaces the Square Meter is the primary unit. Each side of a square meter is 10 decimeters in length, and hence a square meter contains 10 X 10, = 100, square deci- meters. Each side of a square decimeter is 10 centimeters i'^^ MEASURES OF SURFACE. 341 length, and lience a square decimeter contains 10 X 10,= 100, square centimeters, etc. Thus it may be seen that while meas- iTi'es of length increase and decrease by a scale of tens, measures of surface increase and decrease by a scale of hundreds. Since the values of imits of surface increase and decrease by a scale of hundreds, it is necessary, in writing numbers denoting surfaces, to allow tw o places for sq. decimeters, two for sq. centir meters, and two for sq. millimeters : thus, 4 sq. meters, 8 sq. decimeters, are written 4.08^'^™, In what places at the right of the decimal point are sq. deci- meters written ? sq. centimeters ? sq. millimeters ? Examples. 1. Express the following numbers in sq. meters and add thpm : 5 sq. decimeters, 87 sq. meters, 26 sq. centimeters, 5.9 sq. meters. A71S. 92.9526^*^'". 2. How many sq. meters are there in a rectangular court 5 meters long and 22 meters wide? Ans. 110*'^'". 3. What is the cost of polishing the surface of a rectangular piece of marble 2 meters 8 decimeters long, and 1 meter 2 decimeters wide, at $2.50 per sq. meter? Ans. $8.40. 049* We have seen that the meter and its subdivisions are used to measure small surfaces ; but to measure surfaces of great extent, as a field, a township, etc., the Are is the primary unit. The Are is a square whose side is 10 meters and whose surface contains 100 square meters. In land measure, centares, ares, and hectares only are used. Note. — The are equals 119.6 square yards, nearly 4 square rods, or about -^Q^ of an acre. The hectare equals about 2-^ acres. Table, 100 cen'tares (''') = 1 are, marked ("). 100 ARES = 1 hectare " l^"). 550, The following table shows the method of writing numbers in land measure, also the relation of the units to the square meter and its subdivisions. 342 THE METRIC SYSTEM. Table. cj (u a 2 a s U W^. .^ 'B i O I "S^ ^ ^ g 03 w a» o . W ^ o S ^ ^ 1 1.01010101 Examples. 1. Express tlie following in ares and add them : 1.3 hectares, 155.5 ares, 43 hectares, 26 centares. Ans. 4585.76^'". 2. Mr. Jenks owned 25 hectares, 32 ares, 16 centares of land, and afterwards bought 36 hectares, 5 ares, 8 centares ; how many ares did he then have? Ans. 6137. 24*^ 3. A had 6 hectares, 7 ares, 9 centares of land, and sold -^^ of it at $54 an are ; how many dollars did he receive for what he sold? Ans, $5960.52. r MEASURES OF VOLUME. S51, In the measurement of solids, the Cubic Meter is the primary unit. Each edge of a cubic meter is 10 decimeters in length, and hence a cubic meter contains 10 X 10 X 10, = 1000, cubic decimeters. Each edge of a cubic decimeter is 10 centimeters in length, and hence a cubic decimeter contains 10 X 10 X 10, = 1000, cubic centimeters, etc. Thus it may be seen that while measures of length increase and decrease by a scale of tens, measures of volume increase and decrease by a scale of thousands. Hence, in writing num- bers denoting volume, three places must be allowed for cu. deci- meters, three for cu. centimeters, and three for cu. millimeters. In what places at the right of the decimal point are cu. deci- meters written? cu. centimeters? cu. millimeters? Examples. 1. Express the following numbers in cu. meters and add them: 58.5 cubic meters, 1.7 decimeters. ^ns. 58.5017*="". MEASURES OF VOLUME. 343 "^ 2. How many cu. meters in a cube whose edge is 2.7 meters? Ans. 19.683«""\ ~ 3. How many cubic meters of air will a room contain whose length is 5.2 meters, whose breadth is 4 meters, and whose height is 35 decimeters? Ans. 72.8"^" «". 552» For measuring firewood, stone, etc., the Stere is the primary unit. Dekasteres and decisteres are also used. The stere is a cubic meter, or 1.308 cubic yards, and is a little more than ^ of a cord. Table. 10 dec'isteres C^') = 1 stere, marked (*). 10 STERES >i/M, = 1 dekastere " (^^^^^), The method of writing numbers in wood measure is the same as that of writing numbers in measures of length. (Art. 545.) Examples. 1. How many steres will a pile of wood contain that is 1 meter long, 1 meter wide, and 1 meter high ? 2 meters high ? 2. What part of a stere will a pile of wood contain that is 1 meter long, 1 meter wide, and 1 decimeter high ? 3. How many steres in a pile of stone that is 1 meter wide, 8.24 meters long, and 4 decimeters high ? Ans. 3.296^ 4. What must be the height of a pile of wood 2.5 meters long and 1 meter wide, to contain a stere ? Ans. 4*^^""'. MEASURES OF CAPACITY. S5S, In measuring liquids, as milk, and dry articles, as beans, barley, and salt, the Liter is the primary unit. The Liter is 1 cubic decimeter, and contains .908 of a quart dry measure, or 1.0567 quarts liquid measure. Table. ^ 10 mil'liliters ('"^) == 1 centiliter, marked C'^). // 10 cen'tiliters ' = 1 deciliter, " (^«="). ^^ 10 dec'iliters = 1 liter, " O- ^ 10 liters =: 1 dekaliter, " (^^^*0- > 10 dek'aliters = 1 hedoliter, " (^0- ^ 10 hec'toliters = 1 kiloliter, " ("). 344 THE METRIC SYSTEM. Note 1. — A kiloliter has the same capacity as a stare, or cu. meter.. Note 2. — A hectoliter equals about 2^ bushels. Numbers denoting capacity are written in the same manner as numbers denoting length. (Art. 545.) Examples. 1. Express the following numbers as liters and add them: 458 centiliters, 82 dekaliters, 765 milliliters. Ans. 825.345^ 2. From a vessel containing 1 hectoliter of oil were drawn 25 liters, 6 centiliters ; how many liters remained? Ans. 74.94'. 3. How many liters of wheat can be put into a bin that is 2 meters long, 1.3 meters wide, and 1.5 meters high ? Ans. 3900\ 4. What must be the length of a bin that is If. meters wide and 1 meter high, to contain 4000 liters of grain? A71S, 2.5"\ MEASURES OF WEIGHT. 0^4. The Gram is the primary unit of weight. The Gram is the weight, in a vacuum, of a cubic centimeter of distilled water at the temperature when it is most dense, which is at 39-^° Fahrenheit. Note 1. — The gram equals 15.432 grains. Note 2. — The new five-cent piece (of 1866) weighs 5 grams. Table. 10 milligrams (™^) = 1 centigram, marked (««°«g). 10 cen'tigrams = 1 decigram, " (•^^"s). 10 dec'igrams = 1 gram, " (s). 10 GRAMS = 1 dekagram, " (^^^'^s). 10 dek'agrams = 1 hectogram, " (^°). 10 hec'tograms = 1 kilogram, " (^^) or (^). 10 kil'ograms = 1 myriagram, " ('"yriag). 10 myr'iagrams = 1 quintal, " (^). 10 quintals • = 1 tonneau, " C). The kilogram, sometimes called the kilo, is considered the unit in weighing gross, heavy articles. The kilogram equals about 2-i pounds avoirdupois, or, more nearly, 2.2046 pounds. The tonneau equals a little more than 2204 pounds. MEASURES OF WEIGHT. 345 Numbers denoting weight are written in the same manner as numbers denoting length. (See Art. 545.) Examples. 1. Express the following numbers as grams and add them: 8.5 dekagrams, 1000 centigrams, 225 decigrams. Ans. 117.5^. 2. Express the following numbers as kilograms and add them : 7.2 hectograms, 8294 grams, 4 quintals. Ans. 409.014''*^. 3. How many papers, each containing ^ a kilogram, may be filled from 32 myriagrams of coffee? Ans. 640 papers. 4. Bought 1 tonneau of coal for $12 ; what is the cost of 1 kilogram of coal at the same rate? Ans. 12 mills. 5. What weight of mercury will a vessel contain whose capacity is 10 cubic centimeters, mercury being 13.5 times as heavy as water. Ans. 135 grams. 6. In 77.2 grams of gold how many cubic centimeters, gold being 19.3 times as heavy as water? Ans. 4*^"^^'^*^. METRIC MEASURES LEGALIZED. BY THE UNITED STATES WITH THEIR EQUIVALENTS NOW IN USE. Note. — Although the equivalents here given are not entirely accurate, they are those which are established by Congress for use in legal proceed- ings, and in the interpretation of contracts, and are sufficiently exact for all practical purposes. Measures of Length. METRIC DENOMINATIONS AND VALUES. EQUIVALENTS IN DENOMINATIONS IN USE. Myriameter . Kilometer . . Hectometer . Dekameter . Meter .... Decimeter . Centimeter . Millimeter . 10,000 m. . 1,000 ra. . 100 m. . 10 m. . Im. . .1 m. . .01m. . .001m. . 6.2137 miles. 0.62137 mile, or 8280 ft. 10 in. 328 ft. 1 in. 393.7 in. 39.37 in. 3.937 in. 0.3937 in. 0.0394 in. 346 THE METRIC SYSTEM. Measures of Surface. METRIC DENOMINATIONS AND VALUES. EQUIVALENTS IN DENOMI- NATIONS IN USE. Hectare . A.TG •• ....... 10,000 sq. m. 100 sq. m. 1 sq. m. 2.471 acres. 119.6 sq. yards. 1550. sq. inches. Centare • • • • Measures of Capacity. METRIC DENOMINATIONS AND VALUES. EQUIVALENTS IN DENOMINATIONS IN USE. Names. No. of Liters. Cubic Measure. Dry Measure. Liquid or Wine Measure. Kiloliter or Stere 1000 1 cu. m. 1.308 cu. yds. 264.17 gal. Hectoliter . . . 100 .1 cu. m. 2 bu. 3.35 pks. 26.417 gal. Dekaliter. . . . 10 10 cu. dm. 9.08 qts. . . . 2.6417 gal. Liter 1 1 cu. dm. 0.908 qt. . . . 1.0567 qts. Deciliter .... .1 .1 cu. dm. 6.1022 cu. in. 0.845 gUl. Centiliter .... .01 10 cu. cm. 0.6102 cu. in. 0.338 fid oz. Milliliter .... .001 .1 cu. cm. 0.061 cu. in. . 0.27 fld dr. Weights. METRIC DENOMINATIONS AND VALUES. EQUIVALENTS IN DENOMINATIONS IN USE. Names. No. of Grams. Weight of what quan- tity of water at max- imum density. Avoirdupois Weight. Millier or Tonneau . Quintal Myriagram Kilogram or Kilo . . Hectogram Dekagram Gram Decigram Centigram Milligram 1,000,000 100,000 10,000 1,000 100 10 1 .1 .01 .001 1 cu. meter . 1 hectoliter . 10 liters . . . 1 liter .... 1 deciliter . . 10 cu. centim. 1 cu. centim. . .1 cu. centim. 10 cu. millim. 1 cu. millim. . 2204.6 pounds. 220.46 «' 22.046 « 2.2046 " 3.5274 ounces. 0.3527 " 15.432 grains. 1.5432 " 0.1543 " 0.0154 " REDUCTION OF NUMBERS. 347 REDUCTION OF NUMBERS IN THE METRIC SYSTEM TO EQUIVALENTS NOW IN USE. 55^, III. Ex. In 5 meters how many inches? how many ^^^^ ' Explanation. Since 1 meter equals 39.37 inches, 5 meters must opekation. equal 5 times 39.37 inches, and 39.37 X5 i/.irtc/v A since 12 inches equall foot, there = 16.40 rV ft., Ans. , , c 4. ^u 12 1^ » must be as many feet as there are times 12 m 5 times 39.37, which is 16.403^,^ times. Am. 16.40y\ft. Examples. 1. In 1 meter, 2 decimeters, how many feet? Ans. 3.937 ft. 2. In 25 millimeters how many inches ? Ans. .985 in. 3. How many inches long is a silkworm that measures 5.2 centimeters in length ? • Ans. 2.047 in.4-. 4. What is the height of a person in feet and inches whose height is 1 meter, 728 millimeters? Ans. 5 ft. 8 in.+. 5. In 21 ares how many square yards? Ans. 2511.6 sq. yds. 6. In a field of 7 hectares, 2 ares, how many acres? Ans. 17.346 acres-}-. 7. In 32 centares how many square yards ?• Ans. 38.272 sq. yds. 8. In 12 steres of wood how many cubic yards? how many • cords? Ans. 15.696 yds. ; 3.31 cords -f-. 9. In 24 kiloliters how many gallons ? Ans, 6340.08 gal. 10. How many gallons of vinegar may be put into a cask containing 8 dekaliters, 2 liters? Ans. 21.66 gals.-[-. 11. In 2 hectoliters how many bushels and pecks? Ans. 5bu. 2.7 pks. 12. In 23 kilograms how many pounds avoirdupois? I ; Ans. 50.7 lbs.-)-. 13. In ^8-kilograms, 7 hectograms, how many pounds avoir- dupois? Ans. 41.22 lbs.-|-. 14. In 27 tonneaux how many tons? Ans. 29.762 tons -4-. 348 THE METRIC SYSTEM. 5Sy7» The following are some of the measures in common use, with their equivalents in measures of the metric system : — An inch = 2.54 centimeters. A cu. yard = .7646 cu. meter. Afoot = .3048 meter. A cord = 8.624 steres. A yard = .1)144 meter. A liquid quart = .9464 liter. A rod = 5.029 meters. A gallon = 3.786 liters. A mile = 1.6093 kilometers. A dry quart =-1.101 liters. A sq. inch = 6.452 sq. centimet's. A peck = 8.811 liters. A sq. foot = .0929 sq. meter. A bushel = 35.24 liters. A sq. yard = .8361 sq. meter. An ounce av. = 28.35 grams. A sq. rod = 25.29 sq. meters. A pound av. = .4536 kilogram. An acre = .4047 hectare. A ton r= .9072 tonneau. A sq. mile == 259 hectares. A grain Troy = .0648 gram. A cu. inch = 16.39 cu. centimet's. An ounce Troy = 31.104 grams. A cu. foot = .02832 cu. meter. A pound Troy = .3732 kilogram. KEDUCTION OF MEASURES NOW IN USE TO EQUIVALENTS IN THE METRIC SYSTEM. 558. III. Ex, In 5 feet how many meters ? ExPLAXATiON. Since 1 foot equals .3048 TTifttRrs. .5 fpftf. Ans. Opekation. .3048 X 5 = 1.524 .3048 meters, 5 feet must equal 5 times .3048 meters, which is 1.524 meters. Ans. 1.524". Examples. 4-. In 2 rods how many meters ? ■%. In 25 sq. yards how many sq. meters ? -3. In 23 acres how many hectares ? ~4. In 8.2 cords how many steres? -h. In 9.2 liquid quarts how many liters? -^. In 28 grains how many grams ? 7. In 3 yards 1 foot how many meters ? -8. In 2 bushels 3 pecks how many liters? 9. In 8 pounds 7 ounces how many kilos? "^^ For a fuller treatment of this subject see Walton's pamphlet edition of *• The Metric System of Weights and Measures." Ans. 10.058"^. Ans. 20.9«